I'm asked to convert this integral:
$$\int^1_0\int_{-\sqrt{1-y^2}}^0\int^{2-x^2-y^2}_0 x\ dz\ dx\ dy $$
to cylindrical coordinates. This is what I calculated for the limits:
$$\int^{2\pi}_0\int^0_{-1}\int^{2-r^2}_0r\cos\theta \ r\ dz\ dr \ d\theta $$
For $z$ I got $2-r^2$ as the upper limit as $r^2=x^2+y^2$.
For $r$ I got $0$ as the upper limit as $r\cos\theta=0$ and so $r=0$.
For the lower limit I got
\begin{align} r\cos\theta & =-\sqrt{1-r^2\sin^2\theta}\\
r^2\cos^2\theta & =1-r^2\sin^2\theta\\
r^2(\cos^2\theta+\sin^2\theta) & = 1\\
r&=\pm1
\end{align}
This is the first part that confuses me. Should the $r$ limits be $[-1,0]$ or $[-1,1]$? I assume its the former because the latter negates all terms in the integrand and just gives zero.
Then, since the $y$ limits are $[0,1]$, I assumed the $\theta$ limits are $[0,2\pi]$, however because the integrand is a $\cos\theta$ expression, integrating it gives a $\sin\theta$ expression and $\sin(0)$ and $\sin(2\pi)=0$, this also just makes the whole integral equate to zero.
I've tried to look at the problem graphically but the $x$ part confuses me. I've never been given a triple integral to convert to cylindrical coordinates in which the $x$ limits are in this form and am not sure how to approach.
Best Answer
Hint
In fact your integral bounds are wrong (except that for $z$). Note that the set $$\{(x,y)\ |\ -\sqrt{1-y^2}< x< 0\ ,\ 0< y< 1\}$$defines a quarter-circle with the following form in polar coordinates$$\left\{(r,\theta)\ | \ 0<r<1\ , \ {\pi\over 2}<\theta<\pi\right\}$$so the integral would become$$\int_{\pi\over 2}^{\pi}\int_{0}^{1}\int_{0}^{2-r^2}r^2\cos \theta dzdrd\theta$$