Question

[Volumes By Cylindrical Shells] Please check my work- did I do this right?

Answer 1

Set up (do not solve) an integral for the volume of the solid obtained by rotating the region bound by the given curves about the specified axis.

Curves:
x = 4 + y^2
x = 2 - y

Axis:
y = -1

My answer (using the method of cylindrical shells):
\[V = \int\limits_{-1}^{2}2\pi(y+1)(2+y-y^2)dy\]

Answer 2

You can see a 1D graph of the area between the two curves here: http://www.wolframalpha.com/input/?i=area+between+the+curves+x+%3D+4-y%5E2+and+x%3D2-y

Please just tell me if I got the right answer. Thanks!

Answer 3

\[\Large (2+y-y^2)\]This appears to be the height of your cylinders... but it doesn't look right... also you posted x= 4 + y^2 here, and x=4-y^2 on wolfram alpha, so i don't know which is correct...

Answer 4

Whoops, sorry, lack of sleep. The correct curve is shown in wolfram alpha, x = 4 - y^2

Answer 5

That's the right height for finding the area between the curves. Does it need to be different for finding volumes?

Answer 6

No, it just wasn't right because you posted the wrong equations here.

So then your height is\[\Large (4 - y^2) - (2 - y) = 2+y-y^2 \] so it looks set up correctly.

Answer 7

The whole thing is correct? The integral I set up?

What's confusing to me is that the area goes below the x-axis into the negative y range. I'm not sure if that changes how to set up a problem like this?

Also, in my book, none of the examples ever have you set up a rotation around a negative axis, so I just assumed that instead of doing (axis - x) for the radius like if you were given a positive axis to rotate around, you'd do (axis + x) if you're given a negative axis to rotate around. Is this the correct understanding?

Answer 8

That doesn't matter. Your integral looks right. Your radius is the y+1, the height of the cylinder is what you have above.

Answer 9

Thanks!

Answer 10

The red line shows y. You need to add 1, since the radius is 1 unit further away.