Volume of a sphere with a cyclinder drilled through it

A ball of radius 13 has a round hole of radius 6 drilled through its center. Find the volume of the resulting solid.

This doesn't seem to be complicated, but as usual, my answers aren't correct.

Volume of the ball is (4/3)*pi*(13^3).

But what's the volume of the hole? Is the hole cylinder or spherical?

Re: Volume of a sphere with a cyclinder drilled through it

Quote:

Originally Posted by

**fgn** It will not be a cylinder. Look at the top/bottom of the thing that you have drilled out. It will not be flat, it will be curved just like the ball.

First look at the circle $\displaystyle y^2 + x^2 = 13^2 $

To me this is a slice of the ball through the center.

Now consider $\displaystyle y = \int_0^6\sqrt{13^2 - x^2}\,dx $

If you rotate this around the y-axis you'll get the volume of the drilled hole.

So the searched volume $\displaystyle V $ will be

$\displaystyle V = \frac{4\pi}{3}13^3 - \pi\int_0^6(13^2 - x^2)\,dx $

The above work is a good attempt, but is wrong....you are using the disk method because you are integrating pi*r^2, but in this case, r is the equation of the x value, which is in terms of y. so you solve the equation of a cirlce for x instead of y and integrate from -10 to 10. but between y=-8 and y=8, the equation of the cirlce will give us the x value of the cirlce and not the cylinder. to get the x value of the cylinder, we have to set up a piecewise function where x=6 on -8>y>8 and the equation for the cirlce SQUARED(because pi*r^2) solved for x in two integrals from -10 to -8 and from 8 to 10. add those two together and then add the integral of 6^2 from -8 to 8 and then multiply by pi but this is tricky because the x value for the hole is a piecewise function and thus is TRICKY to integrate. so just stick to the curves instead of the lines :) lets use the cylinder method instead of the disk/washer method.

R=13

r=6

here is the correct solution:

$\displaystyle equation of a cirlce = x^2 + y^2 = R^2$ ****NOTE THIS IS BIG R AND NOT LITTLE R******

since the cylinder's radius runs along the y-axis, we have to solve for y (this is the general rule) and we integrate with respect to x since y is in terms of x (aka y = an equation with x in it)

more simply, the range of the integral is in x and the equation of the cylinder is solved to get only x values

since we integrate the cylinder, we use the surface area formula and not the volume formula (integrating moves us one step up and the steps going up are point, surface area, volume, holy holy compliccated)

so integrating surface area gives us the volume

surface area of a cylinder is:

$\displaystyle SA(cylinder)=2*pi*r*h$

but whats r and whats h?

r is 6 right? NO, r is only 6 when we are at the maximum, the radius of the cylinder is just x because the radius runs along the y axis remember? draw a picture of the cylinder on a graph

$\displaystyle r=x$

h is the height of the cylinder, so how do we find that out? refer to a picture and you will see that the only equation we have is that of the circle. this equation can be solved for y and gives us this:

$\displaystyle y=\sqrt{R^2-x^2}$

but this only gives us the y-coordinate, we want the height of the cylinder and all we have is the y value. this y value is actually the radius of the cirlce, which is half the height of the cylinder!(you must see this)

so h is just 2 times y which is $\displaystyle h=2*\sqrt{R^2-x^2}$

so create the integral for your cylinder via x values and you get this:

$\displaystyle 2*pi*\int_0^6\(r*h)dr$ which is plugged in as $\displaystyle 4*pi*\int_0^6\(x*\sqrt{R^2-x^2}\)dx$ you can use trig sub to solve this integral with out a calculator since R is a constant

then just subtract this from the $\displaystyle (4/3)*pi*R^2$

Re: Volume of a sphere with a cyclinder drilled through it

yes thank you for pointing out that a post from 7.5 years ago is incorrect.

Fgn and Thedoge were getting awfully hungry sitting at their pc waiting for an answer.