# Volume bounded by 3 surfaces, did I do this correctly?

1. Aug 1, 2015

### mistanervous

1. The problem statement, all variables and given/known data
Find the volume of the solid bounded by the surfaces
$(x^2 + y^2 + y)^2 = x^2 + y^2$
$x + y + z = 3$
and $z = 0$

2. Relevant equations

3. The attempt at a solution
I begin by converting to polar coordinates to do a cylindrical integration with 3 variables.

$(x^2 + y^2 + y)^2 = x^2 + y^2$ becomes $r^4 + 2r^3\sin\theta + r^2\sin\theta = r^2$

$x + y + z = 3$ becomes $z = 3 - r\cos\theta - r\sin\theta$

I graphed the first equation and found that 2 is the maximum value of r. Now I write a triple integral in cylindrical coordinates.

$$V = \iiint_E f(r\cos\theta,r\sin\theta,z) r\,dz\,dr\,d\theta$$

then
$$\int_{0}^{2\pi}\int_{0}^{2}\int_{0}^{3-r\cos\theta-r\sin\theta}r\,dzdrd\theta$$
then
$$\int_{0}^{2\pi}\int_{0}^{2}3r-r^2\cos\theta-r^2\sin\theta\,drd\theta$$
then
$$\int_{0}^{2\pi}6-\frac{8}{3} \cos\theta-\frac{8}{3} \sin\theta\,d\theta$$
which then becomes
$$[12\pi + \frac{8}{3}] - [-\frac{8}{3}]$$
$$12\pi + \frac{16}{3}$$

I just want to know if what I did makes sense and if this is the correct answer.

2. Aug 2, 2015

### jbstemp

You need to take into account that r depends on theta, and as such you can't simply integrate it from 0 to 2 like you would with a cylinder of constant radius.

After a bit of algebra you should find that r is bounded between 0 and 1-sinθ, so the integral you want to evaluate is:

$$\int_0^{2\pi} \int _0^{1-sin\theta} \int_0^{3-rcos\theta-rsin\theta} rdzdrd\theta$$

which I found to be 23/4 pi, but you may want to double check that for yourself.

3. Aug 2, 2015

### mistanervous

Ah, thank you so much!! I knew it felt too easy

4. Aug 2, 2015

### mistanervous

Can you help me understand how you found that r is bounded by 1-sinθ and 0?

5. Aug 2, 2015

### jbstemp

So you have (x^2+y^2+y)^2 = x^2+y^2.

After converting to cylindrical coordinate and a bit of rearranging you have
$$r^2 + r(sin\theta-1) = 0$$

Solving for r using the quadratic formula will give you r = 0 and r = 1-sinθ.

6. Aug 3, 2015

### LCKurtz

Surely, you would just factor out an $r$ giving $r(r+(\sin\theta-1))=0$ instead of using the quadratic formula.

But that isn't the whole story. What you actually get upon changing to polar coordinates is$$r^4 + 2r^3\sin\theta+\sin^2\theta = r^2$$Upon dividing out $r^2$ and losing the trivial $r=0$:$$r^2+2r\sin\theta + \sin^2\theta = 1$$ $$(r+\sin\theta)^2-1=0$$ $$(r+\sin\theta + 1)(r+\sin\theta -1) = 0$$So you actually get two solutions: $r = \pm 1 - \sin\theta$. It turns out they both give the same cardioid but traced differently.