What is the volume of a solid formed by rotating a region around the y-axis?

[Chas3down]

Homework Statement

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.

Homework Equations

pi r^2

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

 

[Mark44]

Homework Statement

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.

Homework Equations

pi r^2

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

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.

 

[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

 

[Mark44]

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.

 

[Chas3down]

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

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