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.

  • $\begingroup$ 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 $(r,\theta)$. $\endgroup$ – Ted Shifrin Mar 25 '19 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$$\left\{(r,\theta)\ | \ 0<r<1\ , \ {\pi\over 2}<\theta<\pi\right\}$$so the integral would become$$\int_{\pi\over 2}^{\pi}\int_{0}^{1}\int_{0}^{2-r^2}r^2\cos \theta dzdrd\theta$$

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