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?
Best Answer
In cylindrical coordinate system, one has $$r^2 = x^2+y^2.$$ Hence, your integral should become: $$\int_0^{2\pi}\int_0^1\int_{r^2}^12zrdzdrd\theta$$