Triple integral conversion to cylindrical coordinates equals zero

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.

• You need to draw a picture of the region in the $xy$-plane. If $y$ goes from $0$ to $1$ and $x$ goes from $-\sqrt{1-y^2}$, what is the region? Now figure out $(r,\theta)$. – Ted Shifrin Mar 25 at 17:01

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$$\left\{(r,\theta)\ | \ 0so the integral would become$$\int_{\pi\over 2}^{\pi}\int_{0}^{1}\int_{0}^{2-r^2}r^2\cos \theta dzdrd\theta$$