Calculate the triple integral $$\iiint_G\sqrt{x^2+y^2+z^2}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z$$ where $G = \left \{ z = x^2+y^2; y = x; x = 1; y = 0; z = 0 \right \}$

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I tried to solve this by switching to spherical coordinates.

I'm not sure if I got the bounds of integration right, but here is what I ended up with.

$$\begin{align} 0 &\leqslant \phi \leqslant \frac{\pi}{4} \\ \text{arccot}\left(\frac{1}{\cos(\phi)}\right) &\leqslant \theta \leqslant \frac{\pi}{2}\phantom{\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_} \\ \frac{1}{\sin(\theta)\cos(\phi)} &\leqslant r \leqslant \frac{\cos(\theta)}{\sin^2(\theta)} \end{align}$$

Integral: $$\int ^{\frac{\pi}{4}}_{0} \mathrm{d}\phi\int ^{\frac{\pi}{2}}_{\text{arccot}\left(\frac{1}{\cos(\phi)}\right)} \mathrm{d}\theta\int ^{\frac{1}{\sin(\theta)\cos(\phi)}}_{\frac{\cos(\theta)}{\sin^2(\theta)}}r^2\sin(\theta)\,\mathrm{d}r$$

Any help is greatly appreciated! Thank you!

  • 1
    $\begingroup$ the domain has cylindrical, not spherical symmetry. How bout you try those coordinates? $\endgroup$ – spaceisdarkgreen Oct 20 '17 at 0:03
  • $\begingroup$ As it stands, your final integrand is missing a factor of $r$; the $r^2$ you currently have comes from the volume element, but you still have to convert the original integrand, $\sqrt{x^2+y^2+z^2}$, to spherical coordinates (which is just $r$). Also, could you show some of your work for how you calculated the bounds of integration? One reason I ask is that since the origin in included in the region of integration, it seems to me that the lower bound of $r$ should be $0$. $\endgroup$ – Robert Howard Jul 31 '18 at 18:41

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