Volume of the solid

[dear2009]

Hey everybody,


I am stuck on this problem

Find the volume of the solid of revolution formed by rotating the finite region bounded by the graphs of about the y-axis.

This is what I did

x^5 = sin [(pi)(x^2) /2]

2 integral sign from x = 0 to x = a (pi) (sin^2 [(pi)(x^2)/2] - [x^5]^2) dx

(2pi integral sign from x = 0 to x = a (pi) (sin^2 [(pi)(x^2)/2] - x^10 )dx

2pi integral sign from x = 0 to x = a (pi) (sin^2 [(pi)(x^2)/2] - [2 pi integral sign from x = 0 to x = a] (x^10) dx

sorry for making this confusing, but could I get any help on this one, I really dont know how to solve this at all

 

[BigGlenntheHeavy]

I'll give you the solution, you find how i set it up.

\(\displaystyle \pi \int_{0}^{1}\bigg[y^{2/5}-\frac{2arcsin(y)}{\pi}\bigg]dy \ = \ 2-\frac{2\pi}{7} \ = \ 1.10240209873.\)

 

[galactus]

You can set this up using SHELLS as well.

\(\displaystyle 2{\pi}\int_{0}^{1}\left[x\left(sin(\frac{\pi}{2}x^{2})-x^{5}\right)\right]dx\)

Whichever method you find easier. It is nice to try both and see if you get the same solution. You should.