Attached is a diagram setting up the problem.
Because the axis of rotation is horizontal and we are using shell method, the radius will be in the y-direction and integration will need to be in terms of "y".
The general formula for shell method is:
[tex]2 \pi \int\limits^a_b {r(y) h(y)} \, dy [/tex]
r(y) is the radius as shown in diagram and is equal to the distance from y=-7 to upper curve x=y^2. This is distance from y=-7 to x-axis plus y-value along curve.
---> r(y) = y+7
h(y) is height of shell represented as x-distance between the 2 curves.
y =x^2 --> x = sqrt(y), x = y^2
----> h(y) = sqrt(y) - y^2
The limits are determined by y-values of intersection of 2 curves.
y^2 = sqrt(y) ---> y^3 -y = 0 -----> y = 0,1
Now we can write the integral:
[tex]V = 2 \pi \int\limits^1_0 {(y+7)(\sqrt{y} -y^2)} \, dy [/tex]
[tex]V =2 \pi \int\limits^1_0{(-y^3-7y^2+y^{3/2}+7y^{1/2})} \, dy [/tex]
[tex]V = 2 \pi |^1_0 (-\frac{1}{4}y^4 -\frac{7}{3}y^3 +\frac{2}{5}y^{5/2}+\frac{14}{3}y^{3/2})[/tex]
[tex]V = 2\pi \frac{149}{60} = \frac{149}{30} \pi[/tex]
