use triple integration to find the volume of the area inside the sphere x2+y2+z2=4z but outside the sphere x2+y2+z2=4. its urgent pls
A start would be to note that:
1. $\displaystyle x^2 + y^2 + z^2 = 4z \Rightarrow z^2 - 4z = -x^2 - y^2 \Rightarrow (z - 2)^2 - 4 = -x^2 - y^2 \Rightarrow (z - 2)^2 = 4 - x^2 - y^2$.
2. The line of intersection is in the plane $\displaystyle 4 = 4z \Rightarrow z = 1$. Therefore the line of intersection is $\displaystyle x^2 + y^2 = 3$.
you should calculate the volume enclosed by the two spheres and then subtract this volume from the volume of the sphere $\displaystyle x^2+y^2+z^2=4z $.to find the volume of the area inside the sphere x2+y2+z2=4z but outside the sphere x2+y2+z2=4.
To get the enclosed volume:
First draw a simple diagram diagram. Looking at it side-one in the xz-plane (y = 0) it's not hard to see that the top surface is $\displaystyle z = \sqrt{4 - x^2 - y^2}$ and the bottom surface is $\displaystyle z = 2 - \sqrt{4 - x^2 - y^2}$. This supplies the z-integration terminals.
The x and y integration terminals define the region in the xy-plane defined by the circle $\displaystyle x^2 + y^2 = 3$.
So integrate the triple integral first with respect to z and then switch to polar coordinates.
You can use the spherical coordinates, it's easy to do for your problem.
$\displaystyle \begin{gathered}\left\{ \begin{gathered}x = r\sin \theta \cos \varphi , \hfill \\y = r\sin \theta \sin \varphi , \hfill \\z = r\cos \theta . \hfill \\ \end{gathered} \right. \hfill \\dxdydz = {r^2}\sin \left( \theta \right)drd\theta d\varphi \hfill \\ \end{gathered} $
Then the volume is $\displaystyle V = \int\limits_0^{2\pi } {d\varphi } \int\limits_0^{{\pi \mathord{\left/{\vphantom {\pi 3}} \right.\kern-\nulldelimiterspace} 3}} {\sin \left( \theta \right)} d\theta \int\limits_2^{4\cos \theta } {{r^2}dr}$.
The answer must be $\displaystyle \frac{{22\pi }}{3}$.