Volumes by shells (1 Viewer)

[asianese]

Either my setup is completely wrong or I'm way too tired. The area from x=0 to x=2 between the curve 4y=x^2 and its directrix is rotated about its directrix. Using shells, show the volume of the solid generated is .





\\ = 4\pi \left \frac{2y^{\frac{3}{2}}{3}+\frac{2y^{\frac{5}{2}}{5} \right]_{0}^{1} \\ = 4\pi(\frac{2}{3}+\frac{2}{5}) \\ = \frac{64\pi}{15}

 

[tohriffic]

Just note that you'll need to split the volumes up into the darker blue area and the lighter blue area (which is simply a cylinder).

To be honest, I'm not sure what you did in your first line of working but it should look something like this:
<img src="http://latex.codecogs.com/gif.latex?\delta V \approx 2 \pi (1+y)(2-x)\delta y" title="\delta V \approx 2 \pi (1+y)(2-x)\delta y" />
(This will give you the volume of the lighter blue area)
If you are using the 'rectangular prism' method, this is according to the picture:

where the thickness is given by delta y.

[Sorry about my massive pictures, I tried resizing them..s:]

If you don't understand me, I can run you through the method! Good luck though!

 

[SpiralFlex]

^ Asianese considered a infinitely small change in delta y, by doing an annulus times the depth. You considered the shell being "unrolled" so you get a rectangular shape.

 

[tohriffic]

^ Asianese considered a infinitely small change in delta y, by doing an annulus times the depth. You considered the shell being "unrolled" so you get a rectangular shape.

Oh okay, I get it now. Thanks Spiralflex!




@Asianese, how did you get from line 2 to 3? (Sorry I'm kinda a noob.)