Hi we have the following problem:
$\iiint x\,dx\,dy\,dz$ limited by the paraboloid of equation $x=4 y^2+4z^2$ and for the plane $x=4$.
We are having difficulty on finding the limits of each integral and how to turn to polar coordinates.
Could you offer any tips? I can provide more details on the comments on what we tried. Thank you.
Normally, polar coordinates are given as $x=r\cos \theta $ and $y=r\sin \theta $, but that does not necessarily have to be. In this case, I think it is wise to use $y=r\cos \theta $ and $z=r\sin \theta $. Then, your limits of integration would be from the paraboloid to the plane (in $dx$), then from zero to the radius of the bounding circle in the $y$-$z$ plane, then from zero to $2\pi$ in $\theta$ to complete the full revolution around the circle.
An idea: make a substitution change $\,x\leftrightarrow z\,$, so that you have the paraboloid $\,z=4x^2+4y^2\,$ and the plane$\,z=4\,$, and now use cylindrical coordinates and the symmetry of the paraboloid around the $\,z-\,$ axis:
$$\iiint z\,dx\,dy\,dz=4\int_0^1dr\int_0^{\pi/2}d\theta\int_{4r^2}^4 4r^3\,dz$$
Please do note that $\,4r^3=4r^2\cdot r=z|J|$, where $|J|$ is the Jacobian of the transformation into cylindrical coordinates.
Added: We can also do the following:
$$\begin{align} \iiint z\,dx\,dy\,dz &= 4\int_0^1dr\int_0^{\pi/2}d\theta\int_{4r^2}^4 zr\,dz\\ & = 2\pi\int_0^1r\left[\frac{1}{2}z^2\right]\Bigg|_{4r^2}^4\,dr \\ & = \pi\int_0^1(16r-16r^5)\,dr \\ & = 16\pi\left(\frac{1}{2}-\frac{1}{6}\right) \\ & = \frac{16\pi}{3} \end{align}$$
which is the right answer according to the book. I still am not sure what went wrong with the first method which I leave here for others to check and comment.
