Use cylindrical coordinates to evaluate $ \int_{-2}^{2} \int_{0}^{\sqrt{4-x^{2}}} \int_{0}^{\sqrt{16-x^{2}-y^{2}}} zdzdydx $.
Attempt to answer:
To use cylindrical coordinates , we at first have to identify the region of integration.
According to me the regions are hemisphere :
$ x^{2}+y^{2}+z^{2}=16, \ -2 \leq x \leq 2 , \ \ 0 \leq y \leq 2 $.
But I am not sure. Any help me finding the region and the cylindrical coordinates ?
$ \int_{-2}^{2} \int_{0}^{\sqrt{4-x^{2}}} \int_{0}^{\sqrt{16-x^{2}-y^{2}}} zdzdydx $.
We know that cylindrical coordinate is defined as $$(r,\theta,z)$$
Notice that the variables are bounded by
$$-2 \leq x \leq 2$$
$$0 \leq y \leq \sqrt{4-x^2}$$
$$0 \leq z \leq \sqrt{16-x^2-y^2} $$
The relation between r and x&y is
$$x^2+y^2=r^2$$
From the given region
$$y^2+x^2=2^2$$
Know that $r \geq0$
So the region for r is $$0 \leq r \leq 2$$
Now $\theta$
We assume the direction is counter clockwise rotation.
Observe that y is always positive then, it is from $$0 \leq \theta \leq \pi$$
This is by inspection!
Also we can work it out. Recall that
$$x=rcos\theta$$
$$x=-2$$
$$-2=2cos\theta$$
$$\theta=\pi$$
$$x=2$$
$$2=2cos\theta$$
$$\theta=0,2\pi$$
We observe that y must always be positive. $\theta$ is defined as the angle measure from the positive x-axis in a anticlockwise direction. For one compete rotation we have $0$ to $2\pi$.
First Quadrant x& y are positive
$0$ to $\frac{\pi}{2}$
Second Quadrant x is negative whereas y is still positive
$\frac{\pi}{2}$ to $\pi$
Third Quadrant x& y are negative
$\pi$ to $\frac{3\pi}{2}$
Fourth Quadrant x is positive y is negative
$\frac{3\pi}{2}$ to $2\pi$
As for z you do nothing!
$$0 \leq z \leq \sqrt{16-r^2} $$
I hope that you can carry on!