1. Shell Method

I have a problem that I have been struggling with and need help:

Use the shell method to determine the volume of a 24-cm diameter sphere with a 10-cm diameter hole drilled through the centerline.

If someone could explain this to me so I can stop pulling my hair out I would be very happy.

2. Originally Posted by jigger1
I have a problem that I have been struggling with and need help:

Use the shell method to determine the volume of a 24-cm diameter sphere with a 10-cm diameter hole drilled through the centerline.

If someone could explain this to me so I can stop pulling my hair out I would be very happy.
For the Shell method around the y-axis the equation is...

$\displaystyle V=2 \pi \int_a^bxf(x)dx$

imagine that the part being drilled out is centered on the y-axis. It will have a radius of 5cm. The radius of the sphere will be 12cm.

So the equation of the circle will be

$\displaystyle x^2+y^2=144$

Solving for y and using the positive root we get an equation of the upper half of the circle.

$\displaystyle y=\sqrt{144-x^2}$

Then using all of this we can set up the integral

$\displaystyle V= 2\pi \int_5^{12}x\sqrt{144-x^2}dx$

Since we used the upper half of the circle we will only get half of the volume so we will need to multiply by 2. Doing so we get..

$\displaystyle V= 4\pi \int_5^{12}x\sqrt{144-x^2}dx$

setting $\displaystyle u=144-x^2$ then $\displaystyle du=-2xdx$

subbing in we get...

$\displaystyle -2\pi\int_{119}^0u^{1/2}du=2\pi \int_0^{119}u^{1/2}=\frac{4\pi}{3}u^{3/2}|_0^{119}=\frac{4\pi}{3}(119)^{3/2}=\frac{476\pi}{3}\sqrt{119}$

3. Thanks Empty Set, I follow you all the way till you sub in at the end. Where are you getting the 119 from? Can you explain more when you are subbing in?

4. Originally Posted by jigger1
Thanks Empty Set, I follow you all the way till you sub in at the end. Where are you getting the 119 from? Can you explain more when you are subbing in?
$\displaystyle u=144-x^2$

So when you change your limits of integration we evaluate the u expression.

So for the upper limit x=12

$\displaystyle u=144-12^2=0$

so the new upper limit is u=0.

for the lower limit x=5

$\displaystyle u=144-5^2=129$

so the new lower limit is 129.

Then we flip over the integral using the property

$\displaystyle \int_a^bf(x)dx=-\int_b^af(x)dx$

I hope this clears it up...

P.s you could just back substitue after integrating and use your original limits of integration. I just think that it is more work than is needed.