# Finding The Volume Of Solid Using Triple Integrals II

1. Feb 27, 2004

### wubie

Hello,

I am still unsure of my ability to evaluate the volume of a solid using triple integrals. Here is my question:

Now I know that the intersection of the two paraboloids is

9 = x^2 + y^2.

But I am unsure how to set up the triple integral. I was thinking of splitting the volume into two halves.

But as I am writing this I am not sure that I have to do that. I would say I could set up the triple integral with the following limits of integration as so:

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, x^2 + y^2 <= z <= 36 - 3x^2 - 3y^2 }

How does this look? First I would integrate with respect to z, then y then x.

I was thinking that I could split the solid into two triple integrals as such:

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

and

E = {(x,y,z) | -3 <= x <= 3, -(9-x^2)^1/2 <= y <= (9-x^2)^1/2, 9 <= z <= 36 - 3x^2 - 3y^2 }

But I don't see a problem with the first way I set up the triple integral. However I haven't tried to integrate it as of yet.

Any thoughts on the problem would be appreciated. Thankyou.

2. Feb 27, 2004

I really wouldn't do this volume in cartesian coordinates. This one's just begging for cylindrical coordinates (it's got rotational symmetry about the z-axis), so, without further ado:

z = x^2 + y^2 = r^2
z = 36 - 3x^2 - 3y^2 = 36 - 3r^2

r^2 = 36 - 3r^2
r = 3
z = 9

So the integral intersects when r = 3 and z = 9. It's not easy to figure out that it'd be real difficult do this in only one integral, so let's split the solid up into two smaller solids and add them.

Volume 1: The solid enclosed by z = r^2 and the plane z = 9.
Volume 2: The solid enclosed by the plane z = 9 and z = 36 - r^2

I'll do Volume 1, and you can have your hand at Volume 2.

$$\int\!\!\!\int\!\!\!\int_{V_1}1\,dV$$

So we have some kind of triple-integral over some volue. Since we're doing this is cylindrical coordinates,

$$dV = r\,dz\,dr\,d\theta$$
with $dz\,dr\,d\theta$ arranged in whichever order we want.

Just because the math works out the easiest, I'll choose the order I wrote. So we have,

$$\int_{\theta_\textrm{low}}^{\theta_\textrm{high}}\!\!\!\int_{r_\textrm{low}(\theta)}^{z_\textrm{high}(\theta)}\!\!\!\int_{z_\textrm{low}(r,\theta)}^{z_\textrm{high}(r,\theta)}r\,dz\,dr\,d\theta$$

For $V_1$,
$$z_\textrm{low}(r,\theta) = r^2$$
$$z_\textrm{high}(r,\theta) = 9$$
$$r_\textrm{low}(\theta) = 0$$
$$r_\textrm{high}(\theta) = 3$$
$$\theta_\textrm{low} = 0$$
$$\theta_\textrm{low} = 2\pi$$

So that gives us:

$$\int_0^{2\pi}\!\!\!\int_0^3\!\!\!\int_{r^2}^9r\,dz\,dr\,d\theta = \frac{81\pi}{2} = V_1$$

I got [itex]V = V_1 + V_2 = 243\pi[/tex] as the final answer.

If you want, I can write up a way to do it in Cartesian coordinates but, as I said, it's not exactly pretty.

3. Feb 28, 2004

### wubie

Yes. I thought so too. However I usually try to set up the integral using cartesian coordinates then I show the conversion to polar coordinates.

I ended up doing the question in polar coordinates but I don't remember my answer.

I also ended up doing the question in one volume instead of two. I used the upper limit of

36 - 3r^2

and a lower limit of

r^2.

However, intuitively, I thought that splitting up the integral into two separate integrals was the way to go. And I think that is what I should have done in the first place.

Thanks again for the help cookiemonster.

Chhers.

4. Feb 28, 2004

### HallsofIvy

Staff Emeritus
I don't see any reason to "break the solid up into two smaller solids and add them". The parabola z= 36- 3x2- 3y2= 36- 3r2 is always above the parabola z= x2+ y2= r2 so that will be the upper and lower limits on the "z" integration. There is complete circular symmetry so the &theta; integration will be from 0 to 2&pi;. Finally, all of the volume lies inside the circle or intersection (which has radius 3) so the r integration will be from 0 to 3.
In cylindrical coordinates the volume is given by
$$\int_{\theta=0}^{2\pi}\int_{r=0}^3\int_{z=r^2}^{36-3r^2}rdzdrd\theta$$.

That reduces very quickly to $$2\pi\int_0^3r(36-3r^2-r^2)dr$$ or
$$2\pi\int_0^3(36r- 4r^3)dr$$

5. Feb 28, 2004

Y'know what, HallsofIvy (and wubie, originally) is absolutely right. No reason to split them up. Silly cookie should pay more attention.

6. Feb 28, 2004

### wubie

Thanks very much to the both of you for your insights. It's good to be aware of different approaches.