1. ## volume of rotation

Using the shell method, find the volume of the solid of revolution obtained by rotating the region bounded by y = x and y = x2 about the x-axis
<a> pi/2
<b>2pi/5
<c>pi/15
<d>2pi/5

2. Originally Posted by harry
Using the shell method, find the volume of the solid of revolution obtained by rotating the region bounded by y = x and y = x2 about the x-axis
<a> pi/2
<b>2pi/5
<c>pi/15
<d>2pi/5
since we are rotating about the x-axis, we have to do it with respect to y.

y = x
=> x = y

y = x^2
=> x = sqrt(y)

what are our limits of integration?

they are where
sqrt(y) = y
=> y = y^2
=> y^2 - y = 0
y(y - 1) = 0
=> y = 0, y = 1 ......these are our limits.

now see the graph below, we have to look at it sideways now.
note that the height of the piece we want to rotate is the height of the higher graph minus the height of the lower graph. so height = sqrt(y) - y

by the shell method
V = int{(circumference)(height)}dy where circumference is 2pi*r, our radius here is just y

so V = int{2pi*y*h}dy
=> V = int{2pi*y*(sqrt(y) - y)}dy
=> V = 2pi*int{y(y^(1/2) - y}dy
=> V = 2pi*int{y^(3/2) - y^2}dy
=> V = 2pi*[(2/5)y^(5/2) - (1/3)y^3] evaluated between 0 and 1, the limits we found above

=> V = 2pi*[2/5 - 1/3]
=> V = 2pi/15

3. Originally Posted by harry
Using the shell method, find the volume of the solid of revolution obtained by rotating the region bounded by y = x and y = x2 about the x-axis
<a> pi/2
<b>2pi/5
<c>pi/15
<d>2pi/5
for practice, try doing this by the disk (washer) method to verify the answer.

washer method says: