Multivariable Calculus: Volume

Question

Trying to figure out the following problem:

Evaluate the integral $\int\int\int_EzdV$, where E lies above the paraboloid $z = x^2+y^2$ and below the plane $z=6y$. Round the result to the nearest hundredth.

Thus far, I have the following:

$\int\int\int_{x^2+y^2}^{6y}zdzdA$

I was thinking of putting the problem in terms of cylindrical coordinates, such that I have:

$\int\int\int_{r^2}^{6r\sin{\theta}}zrdzdrd\theta$

Just not sure about the limits for r and $\theta$. When I graph the function, I more a less get an idea. $\theta$ would be from 0 to $\pi$ but I am not sure about r. I would assume you equal $x^2+y^2=6y$ and solve for something. But I am not sure. That being said and I put it in cylindrical coordinates. I get $r^2=6r\sin{\theta}$. Solving for $r$ I get, $r$ = $6\sin{\theta}$. Not sure if this is going about it the right way though.

Thanks in advanced.

Answer 1

This isn't the best picture, but hopefully it will orient you as to the geometry of the situation:

Then

$$ I=\iiint_E z\,dV=\int_0^{2\pi}\int_0^3 \left(\int_{x^2+y^2}^{6y} z\,dz\right)\, r\,dr\,d\theta $$ and switching to polar coordinates via $x=r\cos\theta$, $y=r\sin\theta$, $$ I=\int_0^{2\pi}\int_0^3\int_{r^2}^{6r\sin\theta} z\,r\,dz\,dr\,d\theta=243\pi. $$ (I did the integration in Mathematica.)

Answer 2

When you equate $6y = x^2 + y^2$ you get the first place of intersection, and you are correct with what you believe $\theta$ to be.

But try...

$\int_{0}^{\pi}\int_{0}^{6sin\theta}\int_{r^2}^{6r\sin{\theta}}zdzrdrd\theta$

r has limits from 0 to $6sin\theta$ because it will go from 0 to the r "solved."

If this is incorrect, i hope something can be gathered from it, but i believe it to be the correct answer. Just evalute it now.