Question

A ball of radius 15 has a round hole of radius 5 drilled through its center. Find the volume of the resulting solid. Hint: The upper half of the ball can be formed by revolving the region bounded by the curves y.

Answer 1

Answer:

The volume of the ball with the drilled hole is:

$\displaystyle\frac{8000\pi\sqrt{2}}{3}$

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:

$x^2+y^2=15^2\to y=\sqrt{225-x^2}$

Then we set the integral for the volume by using shell method:

$\displaystyle\int_5^{15}2\pi x\sqrt{225-x^2}dx$

That can be solved by substitution:

$u=225-x^2\to du=-2xdx$

The limits of integration also change:

For x=5: $u=225-5^2=200$

For x=15: $u=225-15^2=0$

So the integral becomes:

$\displaystyle -\int_{200}^{0}\pi \sqrt{u}du$

If we flip the limits we also get rid of the minus in front, and writing the root as an exponent we get:

$\displaystyle \int_{0}^{200}\pi u^{1/2}du$

Then applying the basic rule we get:

$\displaystyle\frac{2\pi}{3}u^{3/2}\Bigg|_0^{200}=\frac{2\pi(200\sqrt{200})}{3}=\frac{400\pi(10)\sqrt{2}}{3}=\frac{4000\pi\sqrt{2}}{3}$

Since that is just half of the solid, we multiply by 2 to get the complete volume:

$\displaystyle\frac{2\cdot4000\pi\sqrt{2}}{3}$

$=\displaystyle\frac{8000\pi\sqrt{2}}{3}$