Find volume of the region bounded by... using cylindrical coordinates.

Question

Find the volume of the region in the first octant bounded by the coordinate planes, the plane $x + y = 4$, and the cylinder $y^2 + 4z^2 = 16$.

Answer is: $8\pi-\frac{32}{3}$

I'm trying to solve this using cylindrical coordinates. I understand how to setup the integral. I've ran it through wolframalpha to get the correct result, but I'm stumped on how to solve it by showing the work. I feel like I'm missing a step on how to simplify the limits of the integral. Thanks!

$\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{4}{\cos\theta + \sin\theta}} \int_{0}^{\sqrt{4-\frac{{(r\sin\theta)}^{2}}{4}}} dzrdrd\theta$

Answer 1

Let $X=2z$, $Y=y$ and $Z=x$. Then use the cylindrical coordinates: $$\mbox{Volume}=\frac{1}{2}\int_{\theta=0}^{\frac{\pi}{2}} \int_{\rho=0}^{4} \int_{Z=0}^{4-\rho\sin(\theta)} dZ\,\rho d\rho\, d\theta=8\pi-\frac{32}{3}.$$

Answer 2

Making a coordinate changes we set $u = x, v = y, w = 2z$, then the Jacobian is $\left|\dfrac{\partial(x,y,z)}{\partial(u,v,w)}\right|= \dfrac{1}{2}$, and the equation in the new coordinates is: $u+v = 4, v^2+w^2 = 16$. Observe that the second equation is now a circle centered at the origin in the $vw$ plane having a radius of $4$ instead of previously an ellipse which causes integration a bit more complicated. Thus, using cylindrical coordinates in this new coordinates, $V = \displaystyle \int \int \int_{\text{Region}(u,v,w)} \left|\dfrac{\partial(x,y,z)}{\partial(u,v,w)}\right|dudvdw = \displaystyle \int_{0}^{\frac{\pi}{2}} \int_{0}^4 \int_{0}^{4-r\sin \theta} \dfrac{1}{2}rdudrd\theta=...= 8\pi-\dfrac{32}{3}$.