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# 1

1.1

Multiple Integrals and Vector Calculus
Double integrals

Let f (x, y) be a real valued function deﬁned over a domain Ω ⊂ IR2 . To start with, let us
assume that Ω be the rectangle R = (a, b) × (c, d). We partition the rectangle with node
points (xk , yk ), where
a = x1 < x2 < ...., xn = b, and c = y1 < y2 < ...., yn = d.
Let Rnm be the small sub-rectangle with above vertices. Now we can deﬁne Upper and
lower Riemann sum as 

U (Pn , f ) =
sup f (x, y)|Rnm |
n,m Rnm

and
L(Pn , f ) = 

n,m

inf f (x, y)|Rnm |

Rnm

where |Rnm | is the area of the rectangle Rnm . Here we may deﬁne the norm of partition Pn 

as Pn  = max |xi − xi−1 |2 + |yi − yi−1 |2 . Then by our understanding of deﬁnite integral
i

we can deﬁne the upper, lower integrals and f (x, y) is integrable if and only if
inf{U (P, f ) : P } = sup{L(P, f ) : P }.
The deﬁnite integral is deﬁned as 

f (x, y)dA = inf{U (P, f ) : P } = sup{L(P, f ) : P }.
Ω 

It can be shown that
Ω

f (x, y)dA = lim S(Pn , f ), where S(Pn , f ) = 
Pn →0 

f (xk , yk )|Rnm |.

n,m

We have the following Fubini’s theorem for rectangle:
Suppose f (x, y) is integrable over R = (a, b) × (c, d), then    

b  d 
d  b
f (x, y)dA =
f (x, y)dy dx =
f (x, y)dx dy
R

a

c

c

a

In case of f (x, y) ≥ 0 we may interpret this as the volume of the solid formed by the
surface z = f (x, y) over the rectangle R. This is precisely the ”sum” of areas of the cross 
d
section A(x) = c f (x, y)dy between x = a and x = b. Since x varies over all of (a, b), this 
b
sum is nothing but the integral a A(x)dx.

1

For any general bounded domain Ω, we can divide the domain into small sub domains
Ωk and consider the upper, lower sum exactly as above by replacing Rnm by Ωk . Then
a function f (x, y) : Ω → IR is integrable if the supremum of lower sums and inﬁmum of
upper sums exist and are equal. We may deﬁne 

f (x, y)dA = lim  

Pn →0

Ω

f (xk , yk )|Ωk |

k

The basic properties of the deﬁnite integral like integrability of f ± g, kf and domain
decomposition theorems holds in this case also.
We call a domain as y-regular if Ω = {(x, y) : a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x)} for some
continuous functions g1 , g2 . Similarly, Ω is x-regular if Ω = {(x, y) : c ≤ y ≤ d, h1 (y) ≤
x ≤ h2 (y)}. A domain is called regular if it is either x-regular or y-regular. Then we have
the following Fubini’s theorem for regular domains.
Theorem 1.1 Let f (x, y) be continuous over Ω.
1. If Ω is y-regular, i.e.,Ω = {(x, y) : a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x)}. Then 
b  

g2 (x)

f (x, y)dA =

f (x, y)dy dx.

R

g1 (x)

a

2. If Ω is x-regular, i.e.,Ω = {(x, y) : c ≤ y ≤ d, h1 (y) ≤ x ≤ h2 (y)}. Then  

d 

h2 (y)

f (x, y)dA =
R

f (x, y)dx dy.

c

h1 (y)

This theorem basically says that if a function is integrable over a domain Ω, then the value
of integral is does not depend on order of integration. That is we can integrate with respect
to x ﬁrst followed by y or vice versa.
(x + y + xy)dA where Ω is the triangle bounded by

Example 1: Evaluate the integral
y = 0, x = 1 and y = x.

Ω

Solution: The triangle is regular in both x and y. The given triangle is
{(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x} = {(x, y) : 0 ≤ y ≤ x, y ≤ x ≤ 1}
2

taking it as y− regular we see that the domain is bounded below by y = g1 (x) = 0 and above by y = g2 (x) = x over x = 0 to x = 1. then we approximate the area of Ω as A ∼ k Ωk = k f (xk . By the deﬁnition of Riemann integral this sum con verges to Ω dA as Pn  → 0. Hence   1  1  15 (x + y + xy)dA = (x + y + xy)dx dy = 24 Ω y=0 x=y  Example 2: Evaluate the integral (2 + 4x)dA where Ω is the domain bounded by Ω y = x and y = x2 . we see that the domain is bounded below y x = h1 (y) = y and above by x = h2 (y) = 1 over y = 0 to 1. this is also equal to  (2 + 4x)dx dx. y) = 1. Solution: The two parabola’s intersect at (0. 1).  2. x = 0 and x + y = 2.1 1. 3 Example: Find the volume of the solid under the paraboloid z = x2 +y 2 over the bounded domain R bounded by y = x. Hence the area is  1   √ 2 x 1 dydx = A= 0 2x2 0 √ 2 (2 x − 2x2 )dx = . Hence. 3 . Solution:    1  x (2 + 4x)dA = Ω (2 + 4x)dy dx y=x2 x=0  1 = 0  (2x + 2x2 − 4x3 )dx = 2/3  1 √ y On the other hand.   x=1  x  (x + y + xy)dA = (x + y + xy)dy dx Ω  x=0 1 (x2 + = 0 y=0 2 x3 15 x + )dx = 2 2 24 Similarly. when f (x. the Ω f (x. y)dA is the volume of the solid bounded above by z = f (x. yk )|Ωk | where f = 1. Whenf (x. y=0 x=y Remark 1. y) and below by Ω. Example: Find the area bounded by y = 2x2 and y 2 = 4x. taking it as x− regular. As discussed in the beginning. 0) and (1.Therefore. y) ≥ 0.

a. x = 1 and y = x. the value of integral does not depend on the order of integration. 0 ≤ y ≤ x}. Therefore. Solution: The points of intersection of y = x. then   1  1  sin x sin x dA = dx dy. But when we consider R to be y-regular. Now by Fubini’s theorem. y) : −2 ≤ x ≤ 1. b > 0. we see that R = {(x.   1  y=2−x  2 2 2 2 (x + y )dA = (x + y )dy dx V = R  0 y=x 1 = 0 3 y + yx2 y=2−x y=x dx = 3  1 0 1 ( ((2 − x)3 − x3 ) + x2 (2 − 2x))dx 3 Example: Find the volume of the solid bounded above by the surface z = x2 and below by the plane region R bounded by the parabola y = 2 − x2 . If we take it as x regular. x ≤ y ≤ 2 − x2 }. Since the can be extended as continuous function over R.  1  2−x2  2 1 x dydx = V = x=−2 −2 y=x x2 (2 − x2 − x)dx Change of order  Consider the evaluation of integral R sinx x dA over the triangle formed by y = 0. by the basic properties of Riemann integral the function is integrable. y ≤ x ≤ 1} and try to evaluate the integral. x R x 0 x=y This is singular integral and diﬃcult to evaluate. As we noted earlier R is regular in x and y. then R = {(x. y) : 0 ≤ y ≤ 1. 1). So R = {(x. y = 2 − x2 are x = −2.  ∞ −ax −bx Example: Evaluate the integral 0 e −e dx. 1. y) : 0 ≤ x ≤ 1. Then the given integral is  1 x  1  sin x sin x dA = dydx = sin xdx = 1 − cos 1 R x 0 y=0 x 0 At times this technique can be used to evaluate some complicated deﬁnite integrals. x Solution: This integral is equivalent to   ∞  b  ∞ −ax e − e−bx −xy dx = e dy dx x 0 0 a 4 . for example.Solution: The domain of integration R is Y -regular bounded above by x + y = 2 and below by y = x with x varying over (0. y = x.

θ values. Solution: The given integral is equivalent to   π  1−cos θ 3ydA = r sin θrdrdθ. Changing the order of integration. R : α ≤ θ ≤ β. R θ=0 r=0 5 . mΔr and α... Then we deﬁne the Riemann sum as  f (rk . θ) is deﬁned over a region R deﬁned using the polar equations. 2Δr. rk + Δr/2 and α + kΔθ.we get Sn =   f (rk .. the required area is  π/2  1−cos θ rdrdθ A =4 θ=0 r=0  π/2  π/2 1 π 2 =4 (1 − 2 cos θ + cos θ)dθ = 2( − 2 + cos2 θdθ) 2 2 0 0  Example: Evaluate 3ydA where R is the region bounded below by x-axis and above R by the cardioid r = 1 − cos θ.. Solution: Since the region is symmetric with respect to x -axis and y-axis. θ) = 0. As Pn  → 0. f2 (r. we get    ∞  b  b  ∞ −xy −xy e dy dx = e dx dy 0 a 0 y=a = ln b a Double integrals in Polar form Suppose we are given a bounded region whose boundaries are given by polar equations. R k Example: Find the area common to the cardioids r = 1 + cos θ and r = 1 − cos θ. α + (k + 1)Δθ.The domain of integration is the inﬁnite strip {(x. Suppose f (r. .. g1 (θ) ≤ r ≤ g2 (θ).. θ)rk ΔrΔθ → f (r. Let ΔA be the polar rectangle with sides rk − Δr/2. θ)rdrdθ.. α + Δθ. Then we divide the region into smaller ”polar rectangles” whose sides have constant r. θ) = 0. say f1 (r.α + m Δθ = β. . Then we divide the r range by Δr. y) : 0 ≤ x ≤ ∞.. a ≤ y ≤ b}. Sn = k The area of small ”polar rectangle” Ak is ΔAk = area of outer sector − area of inner sector = rk ΔrΔθ. θk )ΔAk .

z) dy dx dz = f (x. 4 0 0 1. ∞ Example: Evaluate I = 2 e−x dx. y. Then we obtain small rectangular cubes over which the function will be approximated by f (xk yk . Now by our understanding of Riemann sums we choose reﬁnement of partitions in such way that maxk |Ωk | → 0. So representing this in polar form we integrate over {(r. z) is integrable over Ω ⊂ IR3 . y.2 Suppose f (x. z)dV = f (x. y. z) be a real valued function deﬁned over a closed and bounded region of space IR3 . we may write  ∞   ∞   ∞ ∞ 2 2 2 −x2 −y 2 I = e dx e dy = e−(x +y ) dA 0 0 0 0 Where the integration is over the ﬁrst quadrant (0. ∞) × (0. Then we obtain the deﬁnite integral as  Ω f (x. Using the Fubini’s theorem. yk . z)dV = lim Sn . y.2 Triple (Volume) integrals Let f (x. zk )|Ωk |. z) dz dy dx = f (x. Now we want to deﬁne the deﬁnite integral of f (x. We form the Riemann sum  f (xk . y. We partition the region by small planes parallel to the coordinate axes. z) dx dy dz z x y z y x       = f (x. y. z) dz dx dy = f (x. In this case again Fubini’s theorem states Theorem 1. zk ). y. z) over such regions. Therefore. Sn = k where |Ωk | is the volume of the small rectangle. then        f (x. n→∞ Evaluation of integrals in three dimensions is done again using Fubini’s theorem. 0 Solution: Recall that 2I = Γ( 12 ). z) dy dz dx y x z y z x To evaluate the triple integrals we follow the following steps: 6 . y. y. y. the above integral becomes  π/2  ∞ π 2 e−r rdrdθ = . ∞). θ) : 0 ≤ r < ∞. For example the solid ball or rectangular box. z) dx dz dy Ω x y z x z y       f (x. 0 ≤ θ ≤ π2 }. y.

y) of R where R is the projection of Ω onto IR2 . Solution: Draw line parallel to z-axis and note that the upper surfaces are: 2z + y = 2 over triangle bounded by x = 0. It is easy to see from the deﬁnition. The Projection of Ω on xy-plane is the solution of 8 − x2 − y 2 = x2 + 3y 2 =⇒ x2 + 2y 2 = 4. Therefore. Identify the upper curve and lower curve of the projection R and limits of integration. 3 −2 = Example: Find the volume of the region bounded by x + z = 1. the volume of Ω is    |Ωk | = lim 1 |Ωk | = 1 dV V = lim k→∞ k→∞ k Ω k Example: Find the volume of the region bounded by the surfaces z = x2 + 3y 2 and z = 8 − x2 − y 2 . y + 2z = 2 in the ﬁrst quadrant. Therefore the limits of x and y are to be determined by R : x2 + 2y 2 = 4. Identify the upper surface and lower surface through which this line passes at most once.1. Hence    8−x2 −y2 V = dzdA R y=x2 +3y 2   √ 2 2 (4−x )/2 = −2  2 −  √ (4−x2 )/2 (8 − 2x2 − 4y 2 )dydx √(4−x2 )/2 4 (8 − x2 )y − y 3 √ 3 −2 y=− (4−x2 )/2 √  √ 4 2 2 = (4 − x2 )3/2 dx = 8π 2.  2  y/2  2−y  1  2x  1−x 2 V = dz dx dy + dz dy dx y=0 x=0 z=0 x=0 7 y=0 z=0 . y = 2x. Draw a line parallel to z axis that passes through the point (x. y = 1y = 2x and z = 1 − x over the triangle bounded by y = 0. the limits of z are from z = x2 + 3y 2 to z = 8 − x2 − y 2 . where Ω is bounded above by the surface Ω 2 2 z = 8 − x − y and below by the surface z = x2 + 3y 2 . x = 1.  Solution: The volume is V = dzdydx. Therefore. 2. 3.

Then the inverse transformation is x = u + v. y)dx dy = f (g(u. v. 0). Instead we can take the transformation .On the other hand. v)|du dv Ω G where J is the Jacobian given by ∂(x. v))|J(u. Then any function of x. we get Example: (Order of integration) Evaluate  4  2  z=0 4  x=0 2  x/2 I= = z=0 x=0 y=0 2 cos(x2 ) √ dy dx dz. 2 y/2 0 Solution: The domain of integration is a parallelogram with vertices (0. Then the relation between the double integral over G and Ω is   f (x. Then u = 2x−y 2 Example: Evaluate the integral I = 1 1 J = 0 2 8 = 2. One has to divide the domain into 3 domains. y) xu xv J= = ∂(u. y may be written as a function of u. x = 2 and x = 2y. z Substitutions in multiple integrals Suppose a domain G in uv-plane is transformed onto a domain Ω of xy-plane by a transformation x = g(u. v) yu yv  4  1+ y2 2x − y dxdy. y = 2v. (1. by ﬁrst drawing the line parallel to x-axis. z x cos(x2 ) √ dx dz = 2 sin 4. v). (3. v = y2 . v). we get   1 2−2z  1−z dx dy dz V = z=0 y=0 x=0 Taking the line parallel to y-axis we get   1 1−x  2−2z dy dz dx V = x=0 z=0 y=0  4  1  2 2 cos(x2 ) √ dx dy dz. z z=0 y=0 x=2y Solution: Note that the projection of Ω onto xyplane is the triangle bounded by y = 0. h(u. So changing the order of integration in x and y. . v). 0). 4) and (2. 4). y = h(u.

y = r sin θ. w)|dV D where G xu xv xw ∂(x. Now by change of variable formula   2 1 I= f (u + v. 2) and (0. v. z)dV = F (u.so we can take transformation u = x + y and v = y − 2x. Solution: Taking the transformation x = r cos θ. then   f (x. y = 0 and x + y = 1. 0  1  1−x Example: Evaluate the integral I = 0 √ 0 0 x + y(y − 2x)2 dA. 2). w).. v. 0). In this case the integrand is complicated. v. 2/3 1/3 Hence  1  √ u I= 0  uv 2 dv du v=−2u 2 2 Example: Evaluate the integral I = R (2−xdA 2 −y 2 )2 over R : x + y ≤ 1. w). v. Hence the Jacobian 1/3 −1/3 J = = 1/3. the given triangle will be transformed into triangle bounded by v = u.Under this transformation. v. The inverse of this transformation is x = u−v and 3 2u+v y = 3 . w) zu zv zw 9 .. sin θ r cos θ By substitution formula. w)|J(u. (1. y = y(u. the parallelogram is transformed into cube with vertices (0. 2v)2dudv = 2udu dv = 2. 2t2 Substitution formula for triple integrals As discussed above suppose a three dimensional domain G is transformed onto a domain D with a transformation x = x(u. 0)(1. z = z(u. v.. we get cos θ −r sin θ J = = r. w).  2π  1 I= 0 r=0 r dr dθ = 2π (2 − r2 )2  2 1 dt = π/2. y. z) J= = yu yv yw . y. ∂(u. Solution: The given domain is the triangle bounded by x = 0. v = −2u and u = 1. Under this transformation.

Under the transformation this points are mapped to (x1 . 0 ≤ xy ≤ 2. v) = g(u. The surface xy = 2 transforms to v = 2. J 0 0 1 Now by substitution formula. 0 ≤ z ≤ 1}. 2. Then the planes x = 1. v + Δv)). v + Δv). θ are polar coordinates of the projection of P on to xy-plane and z is the z distance of 10 . y) we get the required formula. v).  Example: Evaluate Ω (x2 y + 3xyz)dV where R = {(x. Then by Taylor’s theorem g(u + Δu. Then the Jacobian J is 1 0 0 1 = y x 0 = x = u. v). z) where r. v + Δv) = g(u. The plane y = 0 transforms to v = 0. h(u. (u + Δu. h(u. Let (u. 2 transforms to u = 1. v = xy and w = z. v)). v). y1 ) = (g(u. y4 ) = (g(u. θ. v) + ∂g Δu + o((Δu)2 ) ∂u ∂g ∂g Δu + Δv + o((Δu)2 ) + o((Δv)2 ) ∂u ∂v Then the area of the ”rectangle” in xy-plane ΔA˜k is g(u + Δu.The main idea of the proof is as follows. = u 1 I= Cylindrical coordinates: A point P in the space (IR3 ) is represented by (r. Let ΔAk be its area element. (u + Δu.  2  2  1 1 (uv + 3vw) dw dv du u v=0 w=0 u=1 2 2 3v (v + )dv du = 2u 1 2 0 3 (2 + )du = 2 + 3 ln 2. v) + ΔA˜k ≈|(x3 − x1 )(y3 − y1 ) − (x3 − x2 )(y3 − y2 )| ≈|J|ΔuΔv + o((Δu)2 ) + o((Δv)2 ) Taking this as the area in the Riemann sum of f (x. v)). v + Δv) be the vertices of the rectangle in the uv−plane. y2 ) = (g(u + Δu. (x2 . v + Δv) and (x4 . (x3 . z) : 1 ≤ x ≤ 2. v). h(u + Δu. v + Δv) and (u. solution: We take the transformation u = x. y3 ) = (g(u + Δu. y.

we see that the limits of z are from 0 to x2 + y 2 and the projection onto xy-plane is the disc:R : x2 + (y − 1)2 ≤ 1.    x2 +y2 V = dzdA R z=0 Now taking the cylindrical coordinates x = r cos θ. y = ρ sin φ sin θ. Therefore. z = ρ cos φ The Jacobian in xρ J = yρ zρ this case is xθ xφ ρ sin φ cos θ −ρ sin φ sin θ ρ cos φ cos θ 0 yθ yφ = ρ sin φ sin θ r cos θ sin φ cos φ 0 −ρ sin φ zθ zφ To ﬁnd limits of integration in spherical coordinates.e. r(r − 2 sin θ) = 0 =⇒ r = 0 to r = 2 sin θ  π  2 sin θ  r2 V = rdz dr dθ θ=0  π r=0  2 sin θ = =4 θ=0 r=0  π 4 z=0 r3 dr dθ sin θdθ = 0 5π 4 Spherical polar coordinates: A point P in the space is represented by (ρ. y = r sin θ.the projection from P . y = r sin θ. 11 = ρ2 sin φ . When the Jacobian is xr J = yr zr we take the transformation x = r cos θ. xθ xz cos θ −r sin θ 0 yθ yz = sin θ r cos θ 0 = r zθ zz 0 0 1 Example: Find the volume of the cylinder x2 + (y − 1)2 = 1 bounded by z = x2 + y 2 and z = 0. Solution: Drawing a line parallel to z axis. From the deﬁnition of φ it is easy to see that z = ρ cos φ and x = ρ sin φ cos θ. So it is not diﬃcult to see that the relation with cartesian coordinates: The projection of P on xy-plane has polar representation: x = r cos θ. φ) where ρ is the distance of P from the origin. Therefore r = ρ sin φ. z = z. φ is the angle made by the ray OP with positive z axis and θ is the angle made by the projection of P (onto xy-plane) with positive x-axis. y = r sin θ where r is the distance of the projected point to origin. z = z we get the projection to be r2 cos2 θ + r2 sin2 θ − 2r sin θ = 0 i.. θ.

Let uk . Now let pˆ be the unit normal to the plane containing Rk and ∇f is the normal to the surface. we get  1  2π  π ρ2  sin φ dφ dθ dρ I= 1 + ρ2 ρ=0 θ=0 φ=0  1 √ ρ2 1 √  dρ = 4π( 2 − ln( 2 + 1)). we see that the limits of ρ are from 0 to 2. Example: Evaluate 2 2 2 1+x +y +z Ω Solution: Going to spherical polar coordinates x = ρ sin φ cos θ.  dV  where Ω is the unit ball x2 + y 2 + z 2 ≤ 1. Solution: Going to cylindrical coordinates. We approximate the area of the surface with the area of the tangent plane ΔPk . =(2π × 2) 2 1 + ρ2 0  Example: Evaluate I = Ω xdV where Ω is the part of the ball x2 + y 2 + z 2 ≤ 4 in the ﬁrst octant. y = ρ sin φ sin θ. zk ) of the surface Δσk . Since the domain is the ball of radius 4. Then the area 12 . x = ρ sin φ cos θ. z) = c. So θ varies from 0 to π/2. Assume that this projection is one-one. Rotate this ray away and towards z-axis to ﬁnd the limits of φ 3. Identify the projection R of the domain on the xy-plane and polar form of R to write the limits of θ. The projection is the circle in the ﬁrst quadrant with radius 2. φ). 2. φ varies from 0 to π/2. Hence. yk . ρ = g2 (θ.  2  π/2  π/2 I= ρ sin φ cos θ (ρ2 sin φ) dρ dθ dφ ρ=0 θ=0 φ=0 π/2  π/2  sin2 φ cos θdφ dθ = π =4 θ=0 φ=0 Surface Area and Surface integrals Consider a surface S deﬁned with f (x.1. y. Now consider the parallelogram with ΔPk and ΔAk as upper and lower planes of the parallelogram. z = ρ cos φ. Let R be its projection on xy-plane. onto. vk be the vectors along the sides of the tangent plane ΔPk . Again since it is cut by the xy-plane below. Let ΔPk be the tangent plane at (xk . φ) where it enters the region and leaves the region. z = ρ cos φ. Let Rk be a small rectangle with area ΔAk and let Δσk be the piece of surface above this rectangle. y = ρ sin φ sin θ. Draw a ray from the origin to ﬁnd the surfaces ρ = g(θ.

of ΔPk is |uk × vk | and uk × vk is the normal vector to ΔPk . This angle can be calculated easily by noting that ∇f and uk × vk are both normals to the tangent plane. y. i. the area of the projection of this tangent plane is |(uk × vk ) · pˆ|(proof of this can be seen in Thomas calculus Appendix 8). ΔAk | cos γk | where γk = angle between (uk × vk ) and pˆ. Then OR = OP cos γ where γ is the angle between OP and OR. z) = z − x2 − y 2 = 0.. Thus ∇f and uk × vk are both normals to the tangent plane ΔPk . the angle between pˆ and uk × vk . y.) ΔPk | cos γk | = ΔAk or ΔPk = So |∇f · pˆ| = |∇f ||ˆ p|| cos γk | Therefore. Surface Area ≈  ΔPk = k This sum converges to  |∇f | ΔAk |∇f · pˆ| k  |∇f | dA ˆ| R |∇f · p where R is the projection of S on to one of the planes and pˆ is the unit normal to the plane of projection.e. pˆ = k.e. Example: Find the surface area of the curved surface of paraboloid z = x2 + y 2 that is cut by the plane z = 2. x2 + y 2 ) : x2 + y 2 ≤ 2} is the ˆ Hence disc R : x2 + y 2 ≤ 2. ΔAk = |(uk × vk ) · pˆ| = |(uk × vk )||ˆ p|| cos(angle between (uk × vk ) and pˆ) In other words. Now imagine the Area of plane is nothing but ”sum” of lengths of lines. Solution: The equation of surface is f (x.. The angle between the plane ΔAk and ΔPk is same as the angle between their normals. Surface Area = ∇f = −2xˆi − 2yˆj + kˆ  |∇f | dA S= ˆ R |∇f · p   = 4x2 + 4y 2 + 1 dA R 13 . i. Clearly this is one-one from xy-plane to IR3 . (This formula is simple in case of straight lines: Let OP be the line from origin and let R be the projection of P on x-axis. Since the plane of projection is xy-plane. So the projection of the surface {(x. From the geometry.

z) be the temperature at (x. z) = x2 + y 2 + z 2 − 2 = 0 and we can ˆ The projection is obtained by solving take the projection onto xy-plane. Let T (x. Then the total temperature may be approximated as    ΔAk |∇f | = dA g(xk .e. Then we can think of integration of g over S. we have a temperature distributed over this surface. We approximate this surface area element with its tangent plane ΔPk . y = r sin θ. y = r sin θ. y. yk . zk )ΔPk = g(xk . yk . z = x2 + y 2 . y. z) |∇f · pˆ| S R If the surface is deﬁned as f = z − h(x. ˆ |∇f · k| S R 14 . ∇f = 2xˆi + 2yˆj + 2z kˆ  |∇f · pˆ| = 2z = 2 2 − x2 − y 2 Therefore. y)) dA. y. y. Solution: The equation of surface is f (x. Then we can calculate the total temperature on S using the Riemann integration. then   |∇f | g(x. y.Going to polar coordinates x = r cos θ. √  2  dA S= 2 − x2 − y 2 R  2π  1 √ √ 2 √ = r dr dθ = 2π(2 − 2). y.. We partition R into small rectangles Ak . As we reﬁne the rectangular partition this ΔPK approximated the ΔSk . Now taking limit n → ∞. Suppose. z)dS = g(x. R = x2 + y 2 = 1. y. zk ) | cos γk | |∇f · pˆ| k k k where pˆ is the unit normal to R or the plane of projection. z)dS = g(x. zk ) g(xk . Let ΔSk be the surface above the ΔAk . using polar coordinates x = r cos θ. we get   |∇f | dA g(x. So pˆ = k. yk . Let R be the projection of S on the plane.  2π  √2 √ 1 + 4r2 r dr dθ = 13π S= 0 r=0 Example: Find the surface of the cap obtained by cutting the hemisphere x2 +y 2 +z 2 =  2 by the cone z = x2 + y 2 . 2 − r2 0 0 Surface Integrals Let g(x. y. i. a surface S is heated up. z) of the surface.  x2 + y 2 + z 2 = 2. z) be a function deﬁned over a surface S. y) = 0. h(x.

v)ˆj + h(u. −1). The variable u and v are parameters and R is the parameter domain. 0 ≤ θ ≤ 2π. v) = f (u. Let x = a sin φ cos θ. onto to R of xy plane. θ) = r cos θ. y. and z = a cos φ. r(r. Let x(r. 0 ≤ z ≤ 1 Here cylindrical coordinates provide everything we need. by the planes x = 0 and x = 1.   zdS = S z R 1 dA = Area(R) = 2 z Parametrizations of Surfaces: Let r(u. z ≥ 0. Examples: 1. −1). v)ˆi + g(u. 0 ≤ φ ≤ π. 1). 1). Solution: f = y 2 + z 2 and this surface can be projected 1-1. φ) = a sin φ cos θ ˆi + a sin φ sin θ ˆj + a cos φ k. This projection is the rectangle with vertices (1. 0 ≤ θ ≤ 2π. (0. (0. θ) = r sin θ z =  x2 + y 2 = r the domain of r. A parametrization of sphere x 2 + y 2 + z 2 = a2 Here we take spherical coordinates. y = a sin φ sin θ. we get ˆ r(θ. z) = z over the surface S cut from the cylinder y 2 +z 2 = 1.Example: Integrate g(x. y(r. 2. . θ is 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π. We assume that r is one-to-one on the interior of R so that S does not cross itself. A parametrization of the cone z=  x2 + y 2 . 0 ≤ θ ≤ 2π. 15 0 ≤ φ ≤ π. So pˆ = kˆ  2 y2 + z2 1 |∇f | = = |∇f · pˆ |2z| z Therefore. θ) = r cos θ ˆi + r sin θ ˆj + r k. (1. So the required parametrization is ˆ 0 ≤ r ≤ 1. The range of r is called the surface S. Taking u = φ and v = θ. v)kˆ be a continuous vector function deﬁned on a plane region R.

where R : a ≤ u ≤ b. v) : R → S. 0 ≤ z ≤ 1 Solution: We found a parametrization of the cone as ˆ 0 ≤ r ≤ 1. Then   b d F dS = F (u. r = 6 sin θ. v)k. v) is called smooth surface if ru and rv are continuous and ru × rv is never zero.3. 0 ≤ z ≤ 5 we take cylindrical coordinates. v) = f (u. 0 ≤ θ ≤ 2π. θ) = r sin θ z = z Substituting in x2 + (y − 3)2 = 9 we get r2 + 6r cos θ = 0. We can ﬁnd that |rr × rθ |  √ r2 cos2 θ + r2 sin2 θ + r2 = 2r Therefore.  2π  1 Surface Area = 0 √ √ 2rdrdθ = π 2 0 Surface integrals: Let F (u. x(r. b. Deﬁnition: (Smooth surface): A parametrized surface r(u. Therefore we have ˆ a≤u≤ The Surface area of smooth surface r(u. 0 ≤ z ≤ 5 r(θ. A parametrization of cylinder x2 + (y − 3)2 = 9. v)ˆj + h(u. ˆ 0 ≤ θπ. θ) = r cos θ ˆi + r sin θ ˆj + r k. The total surface area is approximately equal to the sum of area of of this parallelograms  S∼ |ru × rv |ΔuΔv u v bb This sum is a Riemann sum of the integral a a |ru × rv | dudv. r(r. v)ˆi + g(u. c ≤ v ≤ d is   b d |ru × rv | dudv S= a c Problem: Find the surface are of the surface of the cone  z = x2 + y 2 . v)|ru × rv |dudv S a c 16 . Surface Area: We approximate the surface area element by the parallelogram on the tangent plane whose sides are determined by the vectors ru Δu and rv Δv. v) be a continuous function deﬁned on the parametrized surface S: r(u. Therefore. c ≤ v ≤ d. θ) = r cos θ. y(r. z) = 3 sin 2θ ˆi + 6 sin2 θ ˆj + z k.

z) = 3 cos θˆi + 3 sin θˆj + z k. k=1 where Δsk is the length of the curve between tk−1 . the region.. z)ˆj + P (x. < tn = b of [a. As the norm of the partition approaches zero. z)kˆ We assume that M. b]. z=0 1. The given integral is equal to   (x + y + z)dS = (3 cos θ + 3 sin θ + z)|Tθ × Tz |dθdz S S  4  4  2π (3 cos θ + 3 sin θ + z)dθdz = 6π zdz = 48π. z)ˆi + N (x.. The surface can be represented as ˆ T (θ. Then we introduce the partition a = t1 < t2 . for example F (x.. y. Problem: Evaluate the surface integral S (x + y + z)dS over the surface of cylinder x2 + y 2 = 9. y. these sum’s approaches .. z = z over the parameter domain {(θ. z) = M (x. the integrals over paths through vector ﬁled plays important role. Solution: Using the cylindrical coordinates: x = 3 cos θ. N and P . Suppose F represents a force throughˆ a ≤ t ≤ b is a smooth curve in out a region in space and let r(t) = x(t)ˆi + y(t)ˆj + z(t)k. 0 ≤ z ≤ 4}. work done in moving an object along a path against a variable force or to ﬁnd work done by a vector ﬁeld in moving an object along a path through the ﬁeld. A vector ﬁeld on a domain in the plane or in the space is a vector valued function f : IR3 → IR3 with components say M. P are continuous functions. z) : 0 ≤ θ ≤ 2π. N. y. y.3 0 θ=0 Line integrals and Green’s theorem In many physical phenomena. For example. 0 ≤ z ≤ 4. Then the work done by F along the curve is approximately n  Fk · Tk Δsk . Then Fk · Tk is the scalar component of F in the director of T at tk . If Fk denotes the value of F at the point on the corresponding to tk on the curve and Tk denotes the curve’s unit tangent vector a this point. y = 3 sin θ. tk . √ Then |Tθ × Tz | = 9 cos2 θ + 9 sin2 θ = 3.

a 17 . → |− r  (t)| we get  b → − → F ·− r  (t)dt.  b  b → − → − dx dy F · T ( )2 + ( )2 dt. F · T ds = dt dt t=a a Now substituting T (t) = − → r  (t) .

18 . . Solution: The tangent along the curve T is dr dt  1 0  F · T ds = 1 0 1 = 0 F· − d→ r dt dt 2 3t2 + t5 − 2t3 dt = . ∂x ∂y R C where the line integral is along the boundary C of R such that R is on the left as we advance on the boundary. then the integral is equal to −1 times: That is. Let F (x. y) = F1 (x. Equivalently. 1) tˆi + t2ˆj + t3 k.3 The Line integral is independent of choice of parametrization. the line integral over any closed curve is zero. If we change the orientation. Deﬁnition: The orientation of a parameterized curve is the direction determined by increasing values of the parameter. 1. In this case.   − F · dr = F · dr.  Q → − → F · d− r = f (Q) − f (P ). Example: Find the work done by F = 3x2ˆi + (2xz − y)ˆj − z kˆ over the curve r(t) = ˆ 0 ≤ t ≤ 1 from origin to (1. C −C → − Deﬁnition: Conservative vector ﬁeld: A verctor ﬁeld F is called conservative vector ﬁeld if the line integral depends only on the end points. → − Example: Show that F is a conservative vector ﬁeld. y)ˆi + F2 (x. Then    → − → ∂F2 ∂F1 − dA = F · d− r. Therefore. then above integral is zero. → − That is if there exists a potential f : IR3 → IR such that F = ∇f . y)ˆj be continuous and has continuous partial derivatives everywhere in some domain containing R. 3 Green’s theorem in the plane Let R be a closed bounded region in IR2 whose boundary C consists of ﬁnitely many → − smooth curves.The following can be shown: Theorem 1. P It is clear that if P = Q. The line integral over a parametrized curve depends on the orientation.

solution: Take x = a cos t. y. y) : x2 + y 2 ≤ 1}\{0}.4 Gauss and Stokes theorems Let S be a smooth surface and we may choose unit normal n ˆ at P of S. Then take F = − 2 2 2 ∂x ∂y x2π+ y  x +y → − → F · d− r = sin2 θ + cos2 θ = 2π.   ∂F2 ∂F1  → − ∂F1 x ˆ ∂F2 y ˆ = and R ∂x − ∂y = 0. Let F (x. Gauss Divergence Theorem Let Ω be a closed. This gives     dA = xdy and dA = − ydx R C R C respectively. y = b sin t. integral C → − → F · d− r = 0 Area of plane region: Using Green’s theorem. The direction of n ˆ is called positive normal direction of S at P . For example the Mobius strip is not orientable. F2 = x and then F= − y. 0 ≤ t ≤ 2π. C     ∂F2 ∂F1 − dA = 9 dA = 9π. For example. We call a smooth surface S orientable surface if the positive normal at P can be continued in a unique and continuous way to the entire surface. the resulting normal upon returning to P is opposite to the original vector at P .Example: Evaluate Solution:   → − → → − F · d− r for F = (y 2 − 7y)ˆi + (2xy + 2x)ˆj and C : x2 + y 2 = 1. bounded region in IR3 whose boundary is a piecewise smooth orientable → − surface S. F2 = 0. Then by above formula   1 2π 1 A= ab cos2 t − (−ab sin2 t) dt = πab (xy  − yx )dt = 2 0 2 1. Choose F1 = 0. By addition we have 1 A = (xdy − ydx) 2 2 2 Example: Area bounded by ellipse xa2 + yb2 = 1. The double integral is the area A of R. ∂x ∂y C R R Remark: The region R is such that the boundary consists of smooth curves. we can write area of a plane region as a line integral over the boundary. But the line i+ 2 j. z) be a continuous function that has continuous partial derivatives in some domain containing Ω. The theorem does not hold true for regions like punctured disc {(x. A normal at a point P of this strip is displaced continuously along a closed curve C. Then   → − F ·n ˆ dS divF dV = Ω S 19 .

An important consequence of Stoke’s theorem is that ﬂux through S1 or S2 is same. S2 be two surfaces having the same boundary curve C. the integration around C is taken in the way that S lies in the left of C. z = x + 2 and F = (x2 + yez )ˆi + (y 2 + ˆ ze2 )ˆj + (z 2 + xey )k.ˆ ndA where ∂Ω is the boundary of the domain inside the Example: Evaluate ∂Ω → − cylinder x2 + y 2 = 1 and between the planes z = 0. Here n ˆ is the direction of your head while moving along the boundary with surface on your left. Let F be a continuous function which has continuous partial derivatives in a domain containing S. Then   → − → → − F · d− r (∇ × F ) · n ˆ dS = C S where n ˆ is a unit normal vector of S and. → − F . Solution: The intersection of cylinder and sphere is the boundary of cylinder on the plane √ ˆ The projection R of S on z = 12. ∇ · F = 2x + 2y + 2z. ∇ × F = −3x2 y 2 kˆ and |∇f ·ˆ p| theorem  − → − → 3 4 (− )x2 y 2 z dA F · dr = 4 z C R  2π  2 (r2 cos2 θ)(r2 sin2 θ)rdrdθ = −8π. Hence by Stoke’s the xy-plane is the disc x2 + y 2 ≤ 2 . Example: Suppose S is a surface of a light bulb over the unit disc x2 + y 2 = 1 oriented → − 2 2 ˆ with outward pointing normal. → − → − Solution: With the given F . 20 . By Divergence theorem  → − F ·n ˆ dS = ∂Ω    x+2 2(x + y + z)dV = 2 (x + y + z)dz dxdy x2 +y 2 ≤1 Ω  z=0 Stokes’s theorem Let S be a piecewise smooth oriented surface with boundary and let boundary C be a simple → − closed curve. depending on n ˆ .  → − − → − r where F = x2 y 3ˆi + ˆj + z kˆ and C The intersection of the Example: Evaluate C F · d→ cylinder x2 + y 2 = 4 and the hemisphere x2 + y 2 + z 2 = 16. =−3 θ=0 r=0 Suppose S1 .  → − Compute S (∇ × F ) · n ˆ dS. it is not diﬃcult to obtain. The unit normal to the surface is n ˆ = 14 (xˆi + yˆj + z k). → − |∇f | = z4 .where n ˆ is the outer unit   normal vector of S. Suppose F = ez −2z xˆi + (sin(xyz) + y + 1)ˆj + ez sin(z 2 )k. z ≥ 0.

Advanced engineering Mathematics. Hence  → − → F · d− r = 0. E. Thomas’ Calculus.Solution: Enough to take any surface with boundary x2 + y 2 = 1. C References 1. Then F on this is F = xˆi + (y + 1)ˆj. So we take the unit → − → − → − disc x2 + y 2 ≤ 1. Chapters 15. z = 0. 16 2.Kreyszig. Then ∇ × F = 0. Chapter 9 21 .
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