Volume generated by revolving curve around axis

[togo]

Homework Statement

Find the volume generated by revolving the regions bounded by the given curves about the x-axis. Use indicated method in each case.
Question 11: y = x^3, y = 8, x = 0
Question 15: x = 4y - y^2 - 3, x = 0

Homework Equations

for question 11: Shells
for Question 15: Shells
Shells:
2pi int xy dy

The Attempt at a Solution

Question 11:
2pi x(x^3) dy
x^4
1/5x^5 = 1/5(4^5)
= 204.8
*2pi = pi409.6

The answer should be 768/7 pi

Question 15:
I guess the goal here is isolate Y. So far:
y(4y - y^2 - 3) = 4y^2 - y^3 - 3y
4(1/3y^3) - 1/4(y^4) - 3(1/2y^2)
Went back and decided Y needs to be isolated but not confident on if that should be done or how so posted it here.

Thanks

 

[Simon Bridge]

I think the answer here is to understand the method of shells rather than just applying the formula.

For 11. for example, $\int x^4dy \neq \frac{1}{5}x^5 + c$ ... that would be the case if you were integrating with respect to x. You must integrate with respect to y.

You can understand this as follows:

The region to be rotated is between the curve $y=x^3$ and the y axis, between y=0 and y=8.
The strategy is to divide the volume into shells thickness dy, with height x depending on the radius y.

The volume of a single shell of radius y and height x(y) like that is $2\pi x(y)ydy$ and the total volume you want is the sum of all these single-shell volumes from y=0 to y=8. So you need x as a function of y.