calculus one volumes of solids of revolutions questions?
hi, i am preparing for a test and i have a few questions for which i do not have the answer to that i would like to compare. thank you in advance for the help!

1) set up, but do not evaluate the following integrals, which will solve for volume
a) y= 5-x^2 and y=4x, rotated around line y=8
b) y= 5-x^2 and y=4x, rotated around line y=-30
c) y= 5-x^2 and y=4x, rotated around line x=7
d) y= 5-x^2 and y=4x, rotated around line x=-10

2) find the volume obtained as the region bounded by y=tan(x), x=pi/6, x=pi/4 is rotated around x-axis.

3)find the volume obtained as the region bounded by y=x and y=x^2 is rotated around x-axis.

4)find the volume obtained as the region bounded by y=sqrt(x-3), x=4, x=7, y=0 is rotated around the y-axis.

5)set up, but do not evaluate the following integrals which will give surface area.
a) x=10-y^2 and the y-axis, is rotated around y=5.
b) x=10-y^2 and the y-axis, is rotated around x=11.
c) y=sin(pix/4) on [4,8] is rotated around the line y=4pi
d) y=sin(pix/4) on [4,8] is rotated around the line y=-4pi

6) find the exact arc length of:
a) y= 3/8(x^4/3-2x^2/3) on [1,8]
b) 24xy=y^4+48 between y=2 and y=4.

7)the base of a solid is a circular disk with diameter 10. find the volume if each cross-section is a square.

8)prove the formula for surface area of a sphere by considering the semi circle y= sqrt(r^2-x^2) rotated around x-axis
1/
a/ the x-intersect between 2 graphs , x^2+4x-5=0>x = -5 or 1
v = pi { integral of [(4x)^2 - (5-x^2)^2] dx , from x=-5 to x=1 }
given: volume of cylinder = 1617 cm3let r be the radius of base and h be the height of coner:h = 2:3∴ r/h = 2/33r = 2hh = 3r/2 …(i)formula: volume of cylinder = πr2h∴ 1617 = (22/7) r2hr2h = 514.5using (i) we have∴ r2 × (3r/2) = 514.53r3 = 1029r3 = 343r = 7 cmh = 21/2 cmtotal surface area of cylinder = 2πr2 + 2πrh= 2 × (22/7) × 72 + 2 × (22/7) × 7 × (21/2)= 308 + 462= 770 cm2therefore total surface area of cylinder = 770 cm2 the given equations arex2 + y2 = 16 …(i)and y2 = 6x …(ii)on solving the equation (i) and (ii),or x2 + 6x = 16or x2 + 6x – 16 = 0or (x + 8)(x – 2) = 0or x = 2 or – 8 is not possible solution∴ when x = 2, y = ± = ± = ±equation (i) is a circle with centre (0, 0) and meets axes at (±4,0), (0,±4)equation (ii) represents a parabola with axis as x - axis.points of intersection are a () and c (2,)these are shown in the graph below:area bounded by the circle and parabola= 2[area (oado) + area (adba)]area of circle = π(r) 2= π(4)2= 16π sq. unitsthus, required areathe area of the circle x2 + y2 = 16 which is exterior the parabola y2 = 6x is given;y-axis, y = cos x and y = sin x, 0 ≤ x ≤ π/2⇒ sin x = cos xrequired area= (√2 − 1) sq.units 18 sq.units explanation: x2 = 12y ∴ 4b = 12 ∴ b = 3 area of triangle = `1/2 xx "ab" xx "os"` = `1/2 xx 4"a" xx "a"` = `1/2(12)(3)` = 18 sq. units let, y = (1 – 2 tan x)(5 + 4 sin x)we have to find dy/dxas we can observe that y is a product of two functions say u and v where,u = (1 – 2tan x) and v = (5 + 4sin x)∴ y = uvas we know that to find the derivative of product of two function we apply product rule of differentiation.by product rule, we have –…equation 1as, u = (1 – 2tan x)∴ ⇒ ⇒ …..equation 2 {∵ }as, v = 5 + 4sin x⇒ ⇒ …equation 3 {∵ }∴ from equation 1, we can find dy/dx∴ using equation 2 & 3, we get –⇒ ⇒ ∵ sin x = tan x cos x , so we get –⇒ hence, a. dimensions of h = [l]dimensions of s = [mt-2]dimensions of ρ = [ml-3]dimensions of r = [l]dimensions of g = [lt-2]checking the r.h.sdimensionally= = [m0l1t0]a. is dimensionally correctb. dimensions of v = [lt-1]dimensions of p = [ml-1t-2]dimensions of ρ = [ml-3]checking the r.h.sdimensionally= b. is dimensionally correctc. dimensions of v = [l3]dimensions of p = [ml-1t-2]dimensions of r4 = [l4]dimensions of t = [t]dimensions of l = [l]dimensions of η = [ml-1t-1]checking the r.h.sdimensionally= =[l3]c. is dimensionally correctd. dimensions of v = [t-1]dimensions of m = [m]dimensions of g = [lt-2]dimensions of l = [l]dimensions of i = [ml2]checking the r.h.sdimensionally= d. is dimensionally correct solution: `1 + omega + omega^2 = 0`   

`z, z omega , z + z omega`   

`z , z omega , z(1 + omega)`   

`z , z omega , - z omega^2 `   

area `/\_ abc= 1/2 xx b xxh `   

`1/2 xx ab xx od`   

`1/2 xx 1 xx sqrt3/2 = sqrt3/4`   

so `1, omega , -omega^2`   

`z, z omega , - z omega^2 `   

area = `|z|^2 xx sqrt3/4 = 4/sqrt3`   

`|z|^2 = 16/3`   

`|z| = 4/ sqrt3`   

answer solution: `abs(veca)=abs(vecb)=abs(vecc)=lamda`   

`veca*vecb=vecb*vecc*vecc*veca=0`   

`abs(veca+vecb+vecc)^2=(veca+vecb+vecc)*(veca+vecb+vecc)`   

`=abs(veca)^2+0+0+abs(vecb)^2+abs(vecc)^2`   

`=lambda^2+lambda^2+lambda^2`   

`=3lambda^2`   

`abs(veca+vecb+vecc)=sqrt3lambda`   

`costheta\_1=(veca*(veca+vecb+vecc))/(abs(veca)abs(veca+vecb+vecc))=lambda^2/(lambdasqrt3lambda)=1/sqrt3=theta\_1=cos^(-1)(1/sqrt3)`   

similarly,   

`theta\_1=theta\_2=theta\_3=cos^(-1)(1/sqrt3)` <p>concepts and reason</p>
<p>the concept used to solve this problem is the formal charge calculation of the oxygen atoms present in the ozone molecule.</p>
<p>fundamentals</p>
<p>a charge that is assigned to an atom, which is present in the molecule, is represented as a formal charge. in two bonded atoms, a formal charge assumes that the electrons are equally shared between them. the formal charge is calculated using the following formula:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>here, \(\mathrm{v}\) is the number of valence electrons, \(\mathrm{n}\) is the number of non-bonding electrons and \(\mathrm{b}\) is the number of bonding electrons.</p>
<p> </p>
<p>the structure is as follows:</p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p><img src="https://img.wizedu.com/images/5af6d1f54822b4dfb8fe89d91ef2011c5fe3020815aa4.png" alt="" width="88" /></p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(\begin{aligned}=& 6-6-\frac{2}{2} \\ &=-1 \end{aligned}\)</p>
<p>explanation | common mistakes | hint for next step</p>
<p>the marked oxygen atom has 6 valence electrons. it contains 6 non-bonding electrons. the marked oxygen atom is bonded with one oxygen atom. therefore, the bonding electron of oxygen is \(2 .\) substitute the values in a formal charge formula to determine the formal charge of oxygen (marked).</p>
<p> </p>
<p>the structure is as follows:</p>
<p><img src="https://img.wizedu.com/images/39493056a8515708b79350d1976e17b35fe30216f0473.png" alt="" width="88" /></p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(=6-2-\frac{6}{2}\)</p>
<p>\(=+1\)</p>
<p>explanation | common mistakes | hint for next step</p>
<p>the oxygen atom in the middle has 6 valence electrons and has 2 non-bonding electrons. the central oxygen atom is bonded with 1 double bond and 1 single bond. so, there is a total of 3 bonds containing 6 electrons (bonding electrons). the formal charge of the middle oxygen atom can be determined by substituting the values in a formal charge formula.</p>
<p> </p>
<p>the structure is as follows:</p>
<p><img src="https://img.wizedu.com/images/b1122ce7723aa4ab1f9ae5c9e457299e5fe3022ca77a5.png" alt="" width="88" /></p>
<p>calculate the formal charge for the circled oxygen in ozone molecule as follows:</p>
<p>formal charge \(=\mathrm{v}-\mathrm{n}-\frac{\mathrm{b}}{2}\)</p>
<p>\(=6-4-\frac{4}{2}\)</p>
<p>\(=0\)</p>
<p>the formal charge of each oxygen atom in \(\mathrm{o}\_{3}\) molecule are labeled as follows:</p>
<p><img src="https://img.wizedu.com/images/1de5ee227194f6d8c7fb9fb0e76160f25fe3024117fde.png" alt="" width="120" /></p>
<p>the oxygen atom present on the right side contains 6 valence electrons which have 2 non-bonded electrons. the marked oxygen is bonded with 1 double bond with a central oxygen atom. so, the number of bonding electrons is \(4 .\) the formal charge of the corner oxygen atom (right corner) can be determined by substituting the values in the formal charge formula.</p>
<p>the formal charge of each oxygen atom in \(\mathrm{o}\_{3}\) molecule are labeled as follows:</p>
<p><img src="https://img.wizedu.com/images/1de5ee227194f6d8c7fb9fb0e76160f25fe3024117fde.png" alt="" width="120" /></p> given the boundaries of the area to be found are,• the curve • the x-axis• x = 0 (y-axis)• x = 4 (a line parallel toy-axis)as per the given boundaries,• the curve,is a curve with vertex at .• x=2 is parallel toy-axis at 2 units away from the y-axis.• x=0 is the y-axis.• the four boundaries of the region to be found are,•point a, where the curve y2 = 6x + 4 and x=0 meet.•point b, where the curve y2 = 6x + 4 and x=2 meet.•point c, where the x-axis and x=2 meet i.e. c(2,0).•point o, or the origin i.e. o(0,0).area of the required region = area of oabc.[using the formula ]the area of the required region