Question

Consider the given curves to do the following:
x=5y^2 - y^3, x=0

Use the method of cylindrical shells to find the volume V of the solid obtained by rotating the region bounded by te given curves about the x-axis. Sketch the region and a typical shell

Answer 1

But if the region is bounded by x=0, then the graph to the left of the y-axis is irrelevant, no?

Answer 2

this is a modified result; just exchange x for y and graph normally

Answer 3

http://www.wolframalpha.com/input/?i=x%3D5y^2+-+y^3

Answer 4

i cant seem to get a descent graph off of it

Answer 5

I'm confused. If I graph this in terms of y, then the region changes, in which case, the shell changes

Answer 6

thats what im trying to determine at the moment ...

Answer 7

http://www.wolframalpha.com/input/?i=inverse+5y^2+-y^3
thats a better view in my opinion

Answer 8

And then there is the question of the limits of integration for this

Answer 9

graphing for x and y and rotating about yaxis is fine

Answer 10

when does 5x^2-x^3 = 0?

Answer 11

5x^2-x^3 = 0

x^2(5-x) = 0 ; x=0 and x=5

Answer 12

\[2\pi\int_{0}^{5}(5x^2-x^3)dx\]

Answer 13

or just as well:

\[2\pi\int_{0}^{5}(5y^2-y^3)dy\]
gets the same results

Answer 14

you have to include the radius y in that

Answer 15

Yes, for sure. I'm just not clear about how to graph this with respect to y and depict the shell

Answer 16

not with shells;

\[\sum(5y^2-y^3)2\pi\Delta y\]
\[\int(5y^2-y^3)2\pi\ dy\]

Answer 17

the integration limits are the radius

Answer 18

No, I have to use shells, have no choice

Answer 19

u are saying then that we dont need to use the radius y?

Answer 20

the radius IS the limits of integration with shells; unless im recalling it wrong

Answer 21

you will not get the same answer if you dont do (y)(5y^2-y^3)

Answer 22

the circumference of each shell is 2pi "r" where r is being integrated from 0 to 5

Answer 23

But I cannot do this with respect to x. It has to be done with respect to y

Answer 24

roughly drawn it should be something,like this: