Good night, i have a serious problem solving this integral.


I make a change of cylindrical coordinates, and when i make the change my integral change:

$\int_{0}^{1}\int_{0}^{\varPi}\int_{0}^{a}zr^{2}dzd\theta dr$ but when i computing the integral, the answer is: $\frac{8}{9}a^{2} $ and my integral $\frac{1}{6}a^{2}\varPi$ please help me!

  • $\begingroup$ @Bvss12: That's very misleading. You corrected the post, then commented as if you hadn't changed it and heropup's comment had no merit. Please don't do that. $\endgroup$ – joriki May 19 '16 at 0:14

The upper bound for $y$ is $$y=\sqrt{2x-x^2}$$ So $$y^2+x^2-2x=r^2-2r\cos\theta=0$$ So that means the limit on $r$ is $r\le2\cos\theta$. And $0\le\theta\le\frac{\pi}2$ because $y\ge0$ and $x\ge0$. Then we have $$\begin{align}\int_0^2\int_0^{\sqrt{2x-x^2}}\int_0^az\sqrt{x^2+y^2}dz\,dy\,dx&=\int_0^{\frac{\pi}2}\int_0^{2\cos\theta}\int_0^azr\,dz\,dr\,d\theta\\ &=\frac12a^2\cdot\frac13(8)\int_0^{\frac{\pi}2}\cos^3\theta\,d\theta\\ &=\frac43a^2\int_0^{\frac{\pi}2}(1-\sin^2\theta)\cos\theta\,d\theta\\ &=\frac43a^2\left[\sin\theta-\frac13\sin^3\theta\right]_0^{\frac{\pi}2}=\frac89a^2\end{align}$$ EDIT: The situation might be a little more clear if we append a plot of the region of integration. Figure 1

  • $\begingroup$ Thanks for you answer, but i dont understand how you can say $0\leq\theta\leq\frac{\varPi}{2}$? $\endgroup$ – Bvss12 May 19 '16 at 0:37
  • 1
    $\begingroup$ Draw a graph of the region of integration and look at the domain of $\theta$. $\endgroup$ – user5713492 May 19 '16 at 0:50
  • $\begingroup$ Thanks man!, but i have a question, i'm very confused because a think the integration limit go to $0\leq\theta\leq\varPi$, because it's a medium of circle. $\endgroup$ – Bvss12 May 19 '16 at 1:07
  • $\begingroup$ Oh, I see what you are trying to do. If you make the substitution $x=1+r\cos\theta$, $y=r\sin\theta$, then you get the limits you were talking about. But in that case the factor $\sqrt{x^2+y^2}=\sqrt{r^2+2r\cos\theta+1}$, not just $r$ and the integral becomes much more challenging. $\endgroup$ – user5713492 May 19 '16 at 2:48

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