Homework Help: 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

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.