Find the volume of the solid obtained by rotation of the reg

[nikchic5]

Find the volume of the solid obtained by rotation of the region bounded by y = x^2 and y^2 = x, about the x-axis. Sketch the region, the solid, and the typical disk or washer.

For the life of me, I cannot figure out how to do these and I am so confused! Thanks so much for any help!

 

[galactus]

What are the limits of integration?. Can you find those?.

To find them, set the given functions equal and solve to see where they intersect. Graph them.

You can use both shells and washers to do this. Try both and see if you get the same result. I will do one and you try the other, OK?.

Here's shells:

\(\displaystyle \L\\2{\pi}\int_{0}^{1}y(\sqrt{y}-y^{2})dy\)

 

[nikchic5]

Hello...Thanks!

I did the washer formula and my final answer was 3pi?10 is that correct? I'm not to sure on the other one...what did you get?

Thanks so much!
Also, are the point of interection (0,1) and (1,1)??

 

[galactus]

That's correct.

Shells I posted above. It gives the same result as washers.

Washers:

\(\displaystyle \L\\{\pi}\int_{0}^{1}(x-x^{4})dx=\frac{3{\pi}}{10}\)

 

[skeeter]

area of region is between y = x² and y = sqrt(x) in quad I.

points of intersection are (0,0) and (1,1)

using washers ...

\(\displaystyle \L V = \pi \int_0^1 (\sqrt{x})^2 - (x^2)^2 dx\)

\(\displaystyle \L V = \frac{3\pi}{10}\)

 

[nikchic5]

One more question...

How exactly does it work out if you use the other formula...not the washer? Thanks so much