# Thread: volume of a sphere with a hole

1. ## volume of a sphere with a hole

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

2. Originally Posted by amiv4
A ball of radius has a round hole of radius drilled through its center. Find the volume of the resulting solid.
this amounts to finding the volume of rotation for the region bounded by the curves $y = \sqrt{144 - x^2}$ and $y = 7$ revolved about the x-axis

or analogously, the region bounded by $y = \sqrt{144 - x^2}$ and $x = 7$ revolved about the y-axis

3. so would u set it up like this

2pi*integral from 0 to 9.74679 of sqrt(144-x^2)-7 dx

4. Originally Posted by amiv4
so would u set it up like this

2pi*integral from 0 to 9.74679 of sqrt(144-x^2)-7 dx
no

first of all, it seems to me like you are using the disk (washer) method with the first suggestion i gave you. if so, please say that and don't leave us guessing.

secondly, if that is what you are doing, are you following the formula?

5. ya i think im using the washer method. and i thought i was following the formula but guess not

6. Originally Posted by amiv4
ya i think im using the washer method. and i thought i was following the formula but guess not
the formula is $V = \pi \int_a^b (R^2 - r^2)~dx$

where $R$ is the outer radius, and $r$ is the inner radius.

7. ya i thought i did that
i just times it by 2 to get rid of the bottom half fo the integral
so it was

2pi*integral from 0 to 9.74679 of sqrt(144-x^2)^2-7^2 dx

8. Originally Posted by amiv4
ya i thought i did that
i just times it by 2 to get rid of the bottom half fo the integral
so it was

2pi*integral from 0 to 9.74679 of sqrt(144-x^2)^2-7^2 dx
yes, but you didn't have the squares before, and keep parentheses around everything after the integral sign. note that squaring just gets rid of the square root in the first term. and let the upper limit be $\sqrt{95}$, just to be exact