Find volume by revolving around...

[thatguy47]

I can't get the right answer for this problem:
Region bounded by y=x^2 y=2-x and x=0
Find the volume by revolving around x=3
The answer is pi(37/6)

 

[Deleted member 4993]

I can't get the right answer for this problem:
Region bounded by y=x^2 y=2-x and x=0
Find the volume by revolving around x=3
The answer is pi(37/6)

Please share your work with us, so that we know where to begin to help you.

 

[thatguy47]

I was trying to do:
fnint((3-squareroot(y))^2-(3-2+y)^2,y,0,2)

 

[skeeter]

I can't get the right answer for this problem:
Region bounded by y=x^2 y=2-x and x=0
Find the volume by revolving around x=3
The answer is pi(37/6)

method of washers w/r to y will require two integral expressions.

recommend the method of cylindrical shells ...

\(\displaystyle V = 2\pi \int_0^1 (3-x)[(2-x) - x^2] \, dx\)

 

[galactus]

When we use shells, we use cross sections that are parallel to the axis about which we are revolving. Since x=3 is vertical, picture the cross sections being stacked up along the x -axis from 0 to 1. Therefore, integrate with respect to x.

\(\displaystyle 2{\pi}\int_{0}^{1}\underbrace{(x-3)}_{\text{revolve about x=3}}(x^{2}-(2-x))dx\)

To use washers is a little trickier