# Finding The Volume Of Solid Using Triple Integrals

1. Feb 26, 2004

### wubie

Hello,

I am having trouble setting up triple integrals to find a volume of a given solid. Here is one of the questions with which I am having trouble.

Now I can see that the projection of the solid on the xy plane is the circle x^2 + y^2 = 9. And I think I can visualize the plane z = y + 3 with respect to the cylinder - it slices the cylinder in half diagonally. But I am not sure how to set up the triple integral.

This is the way I would set up the integral

Let E be the solid in question. Then

E = {(x,y,z) | -3 <= x <= 3, -(9-x^2)^1/2 <= y <= (9-x^2)^1/2, 0 <= z <= y + 3 }

If I was to integrate using these limits I would then integrate with respect to z first, then y, then x.

Now I know that it would be easier to eventually convert to polar coordinates but I would like to know if the way I set up the triple integral to find the volume of the solid is correct so far.

int.[-3,3] int.[-(9-x^2)^1/2,(9-x^2)^1/2] int.[0,y+3] 1*dz*dy*dx

How does that look so far?

2. Feb 26, 2004

### Tom Mattson

Staff Emeritus
I get the same thing.

3. Feb 26, 2004

### wubie

Thanks Tom.

4. Feb 26, 2004

### wubie

I am still having trouble with this question. The easiest part should be integrating the triple integral but I can't seem to get it.

I first integrate with respect to z.

int.[0,y+3] 1*dz = y + 3.

I now integrate with respect to y.

int.[-(9-x^2)^1/2,(9-x^2)^1/2] y + 3 dy =

(y^2)/2 - 3y | .[-(9-x^2)^1/2,(9-x^2)^1/2] =

(((9-x^2)^1/2)^2)/2 + 3((9-x^2)^1/2)) - [((-(9-x^2)^1/2)^2)/2 + 3(-(9-x^2)^1/2)) =

(9-x^2)/2 + 3(9-x^2)^1/2 - [(9-x^2)/2 - 3(9-x^2)^1/2] = 6(9-x^2)^1/2

Now I should integrate with respect to x. But I can't seem to do it / remember how to integrate this part:

int.[-3,3] 6(9-x^2)^1/2 dx

Any further help is appreciated.

BTW, would I be able to integrate this integral with trig substitution? If so. I cannot remember how to do that. Could someone guide me / help me recall the process? Thanks again.

Last edited by a moderator: Feb 26, 2004
5. Feb 26, 2004

### wubie

Right. It's been awhile since I have done trig. substitution, but I believe that that is the way to go here. (Stop me if I am wrong please). 8)

So I presently have

int.[-3,3] 6(9-x^2)^1/2 dx

I now do my trig. sub. by letting x be 3 sin theta. Therefore

dx = 3 cos theta dtheta

I also change my limits.

If x = 3 then theta should be pi/2 and if x = -3 then theta should be -pi/2.

Now to integrate:

If I let x = 3 sin theta then

int.[-3,3] 6(9-x^2)^1/2 dx

becomes

6 * int. [-pi/2, pi/2] (9 - (3sin theta)^2)^1/2 * 3 cos theta dtheta

which then becomes

6 * int. [-pi/2, pi/2] (9 - 9 sin^2theta)^1/2 * 3 cos theta dtheta

which becomes

6 * int. [-pi/2, pi/2] (9 (1 - sin^2theta))^1/2 * 3 cos theta dtheta

which becomes

6 * int. [-pi/2, pi/2] 3 * (1 - sin^2theta)^1/2 * 3 cos theta dtheta

which then becomes

54 * int. [-pi/2, pi/2] (cos^2theta)^1/2 * cos theta dtheta

then

54 * int. [-pi/2, pi/2] cos theta * cos theta dtheta

which becomes

54 * int. [-pi/2, pi/2] cos^2theta dtheta.

Now using the trig. identity

cos^2 = (1+cos2theta)/2
I get

54 * int. [-pi/2, pi/2] (1+cos2theta)/2 dtheta.

which becomes

27 * [ theta + 1/2 * sin2theta] with the limits [-pi/2, pi/2].

Then sub. in the limits.

27 * [ pi/2 + 0 - (-pi/2 + 0) ] = 27 * 2pi/2 = 27 pi.

Anyway, that is what I did and that is what I got. Have I made errors? Or did I do fine? And is this the volume of the solid?