# Triple integral using spherical coordinates ($\theta$ variation)

I have to calculate the integral $$\int_{-1}^{1}\int_{0}^{\sqrt{1-x^{2}}}\int_{0}^{\sqrt{1-x^{2}-y^{2}}}e^{(x^{2}+y^{2}+z^{2})^{3/2}}dzdydx$$ using spherical coordinates. Since I did: $$\left\{\begin{array}{l} x=\rho\sin\phi\cos\theta \\ y=\rho\sin\phi\sin\theta \\ z=\rho\cos\phi\end{array}\right.$$ I got then: $$0\leq\phi\leq\frac{\pi}{2},\quad0\leq\theta\leq\pi,\quad0\leq\rho\leq1$$ Well, I calculated the integral and I got $$\frac{\pi}{3}(e-1)$$ as answer. But in the book, the correct answer is $$\frac{2\pi}{3}(e-1)$$ And I just got this answer when I change the variation of $\theta$: not from $0$ to $\pi$, but $$-\pi\leq\theta\leq\pi$$ For me, $0\leq\theta\leq\pi$ is the right variation of $\theta$. Why am I wrong?

• I obtained the same $\frac{\pi}{3}(e-1) \approx 1.79938$ as your original result. Either (1) there's a human error in the book, or (2) there's an error in how you setup the integral. Are you sure the region is over a quarter of a sphere and not a hemisphere? Just saying. – Lee David Chung Lin May 11 '18 at 5:07
• Yes, I am sure. – mvfs314 May 11 '18 at 13:28
• Your textbook probably miscalculated this. – poyea May 12 '18 at 20:18