1. ## shell method again?

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

x+y=2
x=3–(y–1)^2

What I did was solve for x for both equations and made them equal to each other and tried to find the points of intersection, but I don't think i'm doing it correctly. Can someone elaborate on what I should do and the steps involved.

I'm not sure on whether to apply the shell or disc method to find the answer as well.

I'm also confused as to when it says, for example, "about the x-axis." Do I solve looking for y or x?

Thank you.

2. Originally Posted by abilitiesz
What I did was solve for x for both equations and made them equal to each other and tried to find the points of intersection, but I don't think i'm doing it correctly.
If you do that before sketching an image of the area to revolve then you won't have much confidence in your result, because you won't be entirely sure what it's for...

Sketch!

3. Originally Posted by tom@ballooncalculus
If you do that before sketching an image of the area to revolve then you won't have much confidence in your result, because you won't be entirely sure what it's for...

Sketch!
When I sketched it, I got it to intersect at (2,0). If that's correct, where do I start now?

4. Originally Posted by abilitiesz
When I sketched it, I got it to intersect at (2,0).
... and where else? And have you imagined it revolving so you can clearly see (imagine) the solid? Then think of a Reimman strip inside the area - horizontal or vertical, it's up to you. But then imagine the strip rotating, first around one axis, then the other. One way you should see a disk or washer, the other way a cylinder. One is contained nicely in the target solid - choose that method.

Of course, if your strip was vertical, you're using y = f(x) in the chosen formula, horizontal, then x = g(y).

5. Originally Posted by abilitiesz
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

x+y=2
x=3–(y–1)^2

$\displaystyle V = 2\pi \int_0^3 y[3-(y-1)^2 - (2-y)] \, dy$

6. Originally Posted by skeeter
$\displaystyle V = 2\pi \int_0^3 y[3-(y-1)^2 - (2-y)] \, dy$
By any chance, can you tell me what program you used to construct the graph? I want to verify if my drawn graph is the actual graph compared to a one done on a computer for later purposes.

And thank you for all the help so far, Tom and Skeeter. I appreciate it.

7. Originally Posted by abilitiesz
By any chance, can you tell me what program you used to construct the graph? I want to verify if my drawn graph is the actual graph compared to a one done on a computer for later purposes.

And thank you for all the help so far, Tom and Skeeter. I appreciate it.

Graph

8. Originally Posted by skeeter

Graph
Ah ok. Thank you! I hope it wasn't in the read here first topics lol. If it was, my apologies. So, I finally calculated the answer, but I seemed to still get it wrong. I computed 51pi for my answer, but it's wrong it says.

I think I'm doing something wrong when I'm deriving the equation that you gave me.

I got y^2/2[3y - y^3 -3y^2 +3y/3] - (2y - y^(2)/2). Then I plugged 0 and 3 into it and got my answer like that. Any input on what i'm doing wrong?

9. Originally Posted by abilitiesz
Ah ok. Thank you! I hope it wasn't in the read here first topics lol. If it was, my apologies. So, I finally calculated the answer, but I seemed to still get it wrong. I computed 51pi for my answer, but it's wrong it says.

I think I'm doing something wrong when I'm deriving the equation that you gave me.

I got y^2/2[3y - y^3 -3y^2 +3y/3] - (2y - y^(2)/2). Then I plugged 0 and 3 into it and got my answer like that. Any input on what i'm doing wrong?
your method of integration is all wrong.

you need to expand the integrand (multiply it all out) , combine like terms, and then find the antiderivative and use the FTC.

10. Ohh nevermind. I figured what I did wrong. Thank you.