What is the 3D region in this triple integral?

Question

$$\int_0^\pi\int_0^{\frac{\pi}{2}}\int_0^{\sin\phi}(\rho^2\sin\phi)d\rho d\phi d\theta $$

The above integral represents the volume of some region in $\Bbb{R}^3$

So , what is the region it is representing in $\Bbb{R}^3$

How do i convert the integral into cartesian coordinates.

Answer 1

That region is what you get if you rotate the half-circle$$\left\{(x,0,z)\in\Bbb R^3\,\middle|\,\left(x-\frac12\right)^2+z^2\leqslant\frac14\text{ and }z\geqslant0\right\}$$around the $z$-axis, but you take only a half-rotation; to be more precise, you take into account only those points $(x,y,z)$ such that $y\geqslant0$. In other words, it's the region below the surface that you see in the picture below and above the plane $z=0$.

That's so because:

• You are told that $0\leqslant\theta\leqslant\pi$. So, you only have points $(x,y,z)$ with $y\geqslant0$.
• You are told that $0\leqslant\phi\leqslant\frac\pi2$. So, you only have points $(x,y,z)$ with $z\geqslant0$.
• You are told that $0\leqslant\rho\leqslant\sin\phi$. This is equivalent to $0\leqslant\rho^2\leqslant\rho\sin\phi$, which, in turn, is equivalent to $x^2+y^2+z^2\leqslant\sqrt{x^2+y^2}$.

It is not hard now to see that this will give you the surface that I described above.