Rotating y=x^(3)+1 about x=-1 Using Washer Method

[hvidales]

Homework Statement

y=x^(3) +1, x=1, y=1; rotated about x=-1

Homework Equations

Washer Method. Pi * Integral from a to b of [Outer radius]^2-[inner radius]^2

The Attempt at a Solution

I understand the shell method version but I wanted to learn the washer way for this one.

Pi* Integral from 1 to 2 of (1+1)^(2)-[1+(y-1)^(1/3)]^(2) dy.
I am sure this is the wrong way to set it up because I get like pi lol. The answer should be 9pi/10. Like I said I know how to do this through the shells but not the washer. I also do not know how one can see that the shells would be easier to use. Thanks for the help!

 

[RoshanBBQ]

You set it up correctly. You must be doing the integration incorrectly.

 

[SammyS]

Homework Statement

y=x^(3) +1, x=1, y=1; rotated about x=-1

Homework Equations

Washer Method. Pi * Integral from a to b of [Outer radius]^2-[inner radius]^2

The Attempt at a Solution

I understand the shell method version but I wanted to learn the washer way for this one.

Pi* Integral from 1 to 2 of (1+1)^(2)-[1+(y-1)^(1/3)]^(2) dy.
I am sure this is the wrong way to set it up because I get like pi lol. The answer should be 9pi/10. Like I said I know how to do this through the shells but not the washer. I also do not know how one can see that the shells would be easier to use. Thanks for the help!

Looks good.

By the way,

$\displaystyle \int_1^2 \left(4-(1+\sqrt[3]{y-1}\,)^2 \right) dy=\frac{9}{10}$​

 

[hvidales]

Thank you both for helping out :). I got it thanks to you guys. Yay! I was just integrating wrong. My mistake was that when I squared the 1/3 for some reason I just removed the root instead of making it 2/3 ha. Once again thank you both :)