I want to calculate triple integral\begin{equation}\int\limits_{-1}^{1}\int\limits_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int\limits_{x^2+y^2}^1 2z dzdydx.\end{equation}(the surface is $z=x^2+y^2$, $0\leq z\leq 1$.)
In MAPLE, I have to calculate it, and the result is$$\dfrac{2}{3}\pi.$$Now I want calculate the triple integral with cylindrical coordinates, become this\begin{equation}\int\limits_{0}^{2\pi}\int\limits_{0}^{1}\int\limits_{r}^1 2zr dzdrd\theta.\end{equation}\begin{eqnarray} \int\limits_{0}^{2\pi}\int\limits_{0}^{1}\int\limits_{r}^1 2zr dzdrd\theta &=& \int\limits_{0}^{2\pi}\int\limits_{0}^{1}\left[z^2r\right]_r^1drd\theta\\ &=& \int\limits_{0}^{2\pi}\int\limits_{0}^{1}\left[r-r^3\right]drd\theta\\ &=& \int\limits_{0}^{2\pi}\left[\dfrac{1}{2}r^2-\dfrac{1}{4}r^4\right]_0^1d\theta\\ &=& \int\limits_{0}^{2\pi}\left[\dfrac{1}{4}\right]d\theta\\ &=&\left[\dfrac{1}{4}\theta\right]_0^{2\pi}\\ &=&\dfrac{1}{2}\pi. \end{eqnarray}The results are different. Am I wrong?
$\endgroup$1 Answer
$\begingroup$In cylindrical coordinate system, one has $$r^2 = x^2+y^2.$$Hence, your integral should become:$$\int_0^{2\pi}\int_0^1\int_{r^2}^12zrdzdrd\theta$$
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