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As I noticed this is half-sphere of radius $r=2$.So I changed the variables to spherical coords:

  • $\int_0^2\!\int_{0}^\frac{\pi}{2}\!\int_0^{2\pi}\!\ (\rho^2sin\phi)(1+\rho\sin\phi\sin\theta\ + \rho\sin\phi\cos\theta)\ d\theta\ d\phi\ d\phi $

Giving me the following:

  • $\int_0^2\!\int_{0}^\frac{\pi}{2}\rho^2sin\phi (2\pi)\ d\phi\ d\rho$
  • $\int_0^2\! \rho^2 (2\pi) (cos\phi \Big|_0^{\phi/2})\ d\rho$
  • $-\int_0^2 \rho^2\ (2\pi) = 8\pi$

but according to my textbook the answers should be either $0$ ,$\frac{16\pi}{3}$ ,$\frac{32\pi}{3}$ or $\frac{4\pi}{3}$

I don´t know if something went wrong doing the integral or changing the variables to polar coords

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    $\begingroup$ $(\cos\phi)' = \color{red}{-}\sin\phi$. and $\int \rho^2 d\rho = \color{red}{\frac13} \rho^3$. $\endgroup$ – achille hui May 28 '19 at 23:16
  • $\begingroup$ @achillehiu I didn´t notice i skipped the $\rho$ integral. Thanks $\endgroup$ – Juan Marin May 28 '19 at 23:22
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I don´t know if something went wrong doing the integral or changing the variables to polar cords

The change of variables is correct (typo withstanding).

$$\int_0^2\!\int_{0}^\frac{\pi}{2}\!\int_0^{2\pi}\!\ (\rho^2\sin\phi)(1+\rho\sin\phi\sin\theta\ + \rho\sin\phi\cos\theta)\ d\theta\ d\phi\ d\color{red}\rho $$

The first evaluation is correct.

$$\int_0^2\!\int_{0}^\frac{\pi}{2} (\rho^2\sin\phi)\ (2\pi)\ d\phi\ d\rho $$

Which equals $$2\pi\cdot\int_0^{\pi/2}\sin\phi~d \phi\cdot\int_0^2 \rho^2~d \rho~$$

Then...

$$2\pi\cdot\left[-\cos\phi\right]_0^{\pi/2}\cdot\left[\tfrac 13 \rho^3\right]_0^2$$

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