# Volume of a rotated solid

1. Mar 26, 2013

### Chas3down

1. The problem statement, all variables and given/known data
Find the volume of the solid formed by revolved the region bound by the graps of y =x^2 + 1, x=0, and y =10 around the y axis.

2. Relevant equations
pi r^2

3. The attempt at a solution

The x bound 0 is given to us.
I solved for y=10 by 10=x^2 + 1
I am not sure if I have to go from -3 to 0, and 0 to 3, or because the bound of 0 was give nto us, does that mean 0 is the lower bound? Also wasn't sure about the rotation around the y axis, I assumed the y axis would be 0 so I didn't factor it into the equation..

3
pi ∫ (x^2 + 1)^2 dx = 348pi/5
0

2. Mar 26, 2013

### Staff: Mentor

This is incorrect. Did you draw a sketch of the region and one of the solid? It looks to me like you are thinking that the region is being revolved around the x-axis, not the y-axis.

This simplest way to do this, I believe, is to use horizontal disks.

3. Mar 26, 2013

### Chas3down

oh, thanks! I watched some videos and looked over my noted, and I think I have a better attempt at it now, I did everything in respects to y...

Limits from 1 to 10
Plugged in x=0 to y=x^2 + 1
y=10 was given to us

solved the equal y=x^2 +1 for x to get x = sqrt(y-1)

End up with:

10
pi∫(y-1)dy
1

4. Mar 26, 2013

### Staff: Mentor

Yes, that's the correct integral now. It's easy to evaluate.

It's important to sketch graphs, both of the region and the solid formed by revolving the region. Many students think they are saving time by skipping this step, but they wind up taking more time because of incorrect answers.

5. Mar 26, 2013

### Chas3down

Alright, thanks for not spoon feeding it, starting to get this volume rotation.

Last edited: Mar 26, 2013