The volume of the solid generated by rotating x = 2y^3 - y^4 around the x-axis with the y-axis is?

Possible answers given are: a) 6(pi)/5 b) 64(pi)/15 c) (pi)/15 d) None of these

I tried finding the values of y and I got 0 and 2. I was planning on using the washer method like: (pi) * (definite integral of (2y^3 - y^4)^2 from 0 to 2). Is this correct?

Question

The volume of the solid generated by rotating x = 2y^3 - y^4 around the x-axis with the y-axis is?

Possible answers given are: a) 6(pi)/5  b) 64(pi)/15  c) (pi)/15  d) None of these

I tried finding the values of y and I got 0 and 2. I was planning on using the washer method like: (pi) * (definite integral of (2y^3 - y^4)^2 from 0 to 2). Is this correct?

arwym · Answer

I would really appreciate the help. I have found nothing like this online and the book didn't help. Our professor told us we had to learn this on our own and I have to hand over a test tomorrow. I have tried the book and many online sources but nothing tells what I am supposed to do in a case like this. I have no way to know if this is the way to solve this problem.

anonymous · Answer

I would use shell method here because you are rotating around x-axis and x is function of y. So to try washer method, you would need y(x) or the inverse which is very hard to get in this case. shell method is adding up circumference areas from inner point to outer rim, sort of like tree rings $$\int\limits_a^b 2 \pi r* h $$ radius is the y-value, height is the x value $$\int\limits_0^2 2 \pi y (2y^3 -y^4) dy$$ here is online reference for these types of problems... http://tutorial.math.lamar.edu/Classes/CalcI/VolumeWithCylinder.aspx