I have the following integral:
$$ \iiint zdxdydz$$
on the area bound by the following surfaces: $$z=\frac{a^2}{b^2} (x^2+y^2)$$
$$z=a$$
$$a,b>0$$
The first surface is a paraboloid and the second is a plane. Now, this is how I proceeded:
Introducing cylindrical coordinates:
$x=r\cos\phi$
$y=r\sin\phi$
$|J| = r$
now, what I did next was substitute $z=a$ in the equation of the paraboloid. I got that $x^2+y^2 = \dfrac{b^2}{a}$, or that $r=\dfrac{b}{\sqrt{a}}$ which led me to believe that $$r\in \left[0, \frac{b}{\sqrt{a}}\right]$$
$$\phi \in \left[0, 2\pi\right]$$
Now regarding $z$, I know that it is bound from below by the paraboloid and from above by the plane, so I deduced that the boundaries are: $$z\in \left[\frac{a^2}{b^2}r^2, a\right]$$.
My integral gets the form:
$$\int_0^{\frac{b}{\sqrt{a}}} \int_0^{2\pi} \int_{r\frac{a^2}{b^2}}^a zr dz d\phi dr$$
Which evaluates to $\dfrac{\pi ab^2}{3}$, but my workbook solution is $\dfrac{a^2 b^2 \pi}{4}$.
Could anyone tell me where I went wrong?
Best Answer
Your question reads over the given area. To avoid any confusion, it should rather read over the given region bound by given surfaces.
The workbook answer is wrong. While it says integrand is $z$ but it gives you answer for integrating $z^2$ over the given region.
I just validate going in the order $dr \ dz \ d\phi$,
$ \displaystyle \int_0^{2\pi} \int_0^{a} \int_0^{\sqrt{b^2 z / a^2}} r \ z \ dr \ dz \ d\phi = \dfrac{\pi a b^2}{3}$
which matches your answer.
Whereas,
$ \displaystyle \int_0^{2\pi} \int_0^{a} \int_0^{\sqrt{b^2 z / a^2}} r \ z^2 \ dr \ dz \ d\phi = \frac{\pi a^2b^2}{4}$