In an integration exercise, I need to find the volume of the region bounded by the paraboloid $z=x^2+y^2$ and below by the triangle enclosed by the lines $y=x$, $x=0$, and $x+y=2$ in the $xy$-plane.

Here is the illustration of $xy$-plane and the standard solution:

enter image description here

However, I got some troubles when I try to use the method of polar coordinate transformation to solve it.

$\left\{ \begin{array}{ll} x=r\cdot cos\theta\\ y=r\cdot sin\theta \end{array} \right.$ , where the Jacobian is $J=r$.

The integral should be

$$ \iint_R[(r\cdot cos\theta)^2+(r\cdot sin\theta)^2]\cdot|J|~drd\theta=\iint_R r^3~drd\theta $$

I think the limits of $\theta$ should be $\frac{\pi}{4}\sim\frac{\pi}{2}$, but I'm not sure what are the inner limits. Thanks for helps.


1 Answer 1


You're doing it right, well done!

For the limits, as you said, $\theta$ goes from $\dfrac{\pi}{4}$ and $\dfrac{\pi}{2}$. For $r$, some drawing helps:enter image description here

Here $\alpha=\theta-\dfrac{\pi}{4}$.

The limits of $r$ is from $0$ to some length that changes according to $\theta$ (or $\alpha$). For the highest value, you see in the drawing that $\cos\alpha=\dfrac{\sqrt{2}}{r}$ so $r=\dfrac{\sqrt{2}}{\cos\alpha}$, and you have $\alpha=\theta-\dfrac{\pi}{4}$...

  • 1
    $\begingroup$ @DarrenTsai It's a pleasure :) Yes you're right! Sorry. $\endgroup$ Jan 20, 2019 at 12:38

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