# Find the volume´s integral $\iiint\limits_E\ (1+x+y) dV$ with the inequalities $x^2+y^2+z^2 \leq4\$ and $z \geq0\$

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

• $(\cos\phi)' = \color{red}{-}\sin\phi$. and $\int \rho^2 d\rho = \color{red}{\frac13} \rho^3$. – achille hui May 28 '19 at 23:16
• @achillehiu I didn´t notice i skipped the $\rho$ integral. Thanks – Juan Marin May 28 '19 at 23:22

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...

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