# Triple Integral ( IS THIS RIGHT?)

1. Nov 8, 2009

### een

Triple Integral ( IS THIS RIGHT??)

1. The problem statement, all variables and given/known data
Let R be the solid enclosed by the planes x=0, y=0, z=2 and the surface x^2+y^2, where
x$$\geq$$0, y$$\geq$$0. Compute$$\int\int\int$$xdxdydz

2. Relevant equations

3. The attempt at a solution
I did $$\int$$0-1$$\int$$0-1$$\int$$(x^2+y^2)-2 xdzdydx

$$\int$$0-1$$\int$$0-12x-x^3-xy^2 dydx
$$\int$$0-1[2xy-x^3y-xy^3/3]0-1 dx
$$\int$$0-12x-x^3-x/3dx = 1-1/4-1/6 = 7/12
Can someone tell me if this is the correct answer

2. Nov 8, 2009

### lanedance

Re: Triple Integral ( IS THIS RIGHT??)

$$\int_{0}^1dx\int_{0}^1dy$$

3. Nov 9, 2009

### kandelabr

Re: Triple Integral ( IS THIS RIGHT??)

you got x and y's upper limits wrong.
you are calculating the volume between these two surfaces (z(x,y) = x2 + y2, z < 2), see attachment.

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Last edited: Nov 9, 2009
4. Nov 9, 2009

### een

Re: Triple Integral ( IS THIS RIGHT??)

okay.. this is what i got...
the lower limit of x is 0 the upper limit is sqrt(2)
same for y
the lower limit of z is x^2+y^2 and the upper limit is 2...
is that right?

5. Nov 10, 2009

### kandelabr

Re: Triple Integral ( IS THIS RIGHT??)

yes, my result is 4/3.

6. Nov 10, 2009

### een

Re: Triple Integral ( IS THIS RIGHT??)

i am getting sqrt(2)/3... what am i doing wrong?... i used the limits that you told me and im pretty sure that i integrated correctly... can you show me step by step

7. Nov 11, 2009

### Staff: Mentor

Re: Triple Integral ( IS THIS RIGHT??)

I have switched the order of integration, leaving the integrand unchanged. I think these are the correct limits of integration.
$$\int_{x = 0}^{\sqrt{2}} \int_{y = 0}^{\sqrt{2 - x^2}} \int_{z = x^2 + y^2}^{2} x~ dz dy dx$$

The projection of this region onto the x-y plane is the disk x^2 + y^2 <= 2, a disk of radius sqrt(2) centered at the origin. Geometrically, we have a stack of blocks dx by dy that extend from the paraboloid up to the plane z = 2. Then we let the stack sweep out along the y-axis from y = 0 to y = sqrt(2 - x^2) (i.e., out to the boundary of the disk), then finally let this wall of blocks sweep from x = 0 to x = sqrt(2).

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