Answer:
The volume of the ball with the drilled hole is:
[tex]\displaystyle\frac{8000\pi\sqrt{2}}{3}[/tex]
Step-by-step explanation:
See attached a sketch of the region that is revolved about the y-axis to produce the upper half of the ball. Notice the function y is the equation of a circle centered at the origin with radius 15:
[tex]x^2+y^2=15^2\to y=\sqrt{225-x^2}[/tex]
Then we set the integral for the volume by using shell method:
[tex]\displaystyle\int_5^{15}2\pi x\sqrt{225-x^2}dx[/tex]
That can be solved by substitution:
[tex]u=225-x^2\to du=-2xdx[/tex]
The limits of integration also change:
For x=5: [tex]u=225-5^2=200[/tex]
For x=15: [tex]u=225-15^2=0[/tex]
So the integral becomes:
[tex]\displaystyle -\int_{200}^{0}\pi \sqrt{u}du[/tex]
If we flip the limits we also get rid of the minus in front, and writing the root as an exponent we get:
[tex]\displaystyle \int_{0}^{200}\pi u^{1/2}du[/tex]
Then applying the basic rule we get:
[tex]\displaystyle\frac{2\pi}{3}u^{3/2}\Bigg|_0^{200}=\frac{2\pi(200\sqrt{200})}{3}=\frac{400\pi(10)\sqrt{2}}{3}=\frac{4000\pi\sqrt{2}}{3}[/tex]
Since that is just half of the solid, we multiply by 2 to get the complete volume:
[tex]\displaystyle\frac{2\cdot4000\pi\sqrt{2}}{3}[/tex]
[tex]=\displaystyle\frac{8000\pi\sqrt{2}}{3}[/tex]
