Determine the volume of a solid in space

Question

I'm trying to determine the volume of the solid defined by the following inequalities

$$ [(x,y,z)\in \mathbb{R^3}: x^2+y^2+z^2 \leq 4 , x \geq 0 , z \leq 1] $$

So I tried to integrate with respect to $z$ first and then used polar coordinates for $xy$:

$$ \iint \left( \int_{-\sqrt{4-x^2-y^2}}^1 1 dz \right) dxdy = \iint 1 - \sqrt{4-x^2-y^2} dxdy =$$ $$ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_{0}^{2} r(1-\sqrt{4-r^2}) dxdy = ... = \frac{14\pi}{3} $$

Which I think is wrong! Is there any other way to solve this problem?

Answer 1

You might want to approach it using spherical coordinates by computing the integral:

$$\int_0^\pi\int_0^{\sin^{-1}(1/2)}\int_0^{1/\cos\phi}\rho^2sin\phi \quad d\rho d\phi d\theta$$

Answer 2

It seems pretty clear to me that the restriction on $x$ just cuts the volume in half, which can be calculated by integrating cylinders

$$V = \frac \pi 2 \int_{-2}^1(4-z^2) dz = \frac \pi 2 \bigg [ (4)(3)-\frac 13(1-(-8)) \bigg] = \frac{9\pi}2$$