Volumes of revolution

[dear2009]

Hey everybody,


Find the volume of the solid of revolution formed by rotating the region bounded by the x-axis and the graph of , from x=0 to x=1, about the y-axis.


This is what I began with:
x^ 9/4
x^(4/9)(9/4)
x = 1
pi (y^9/4)^2 = pi y ^ 18/4
18/4 + 1 / 18/4 + 1
= 4/22pi(1^(18/4)
=

after this I dont know what to do. Any help would be greatly appreciated

 

[galactus]

Hey everybody,


Find the volume of the solid of revolution formed by rotating the region bounded by the x-axis and the graph of right here. what is your function?. Is it x^(4/9)? , from x=0 to x=1, about the y-axis.


This is what I began with:
x^ 9/4
x^(4/9)(9/4)
x = 1
pi (y^9/4)^2 = pi y ^ 18/4
18/4 + 1 / 18/4 + 1
= 4/22pi(1^(18/4)
=

after this I dont know what to do. Any help would be greatly appreciated

Going with that, since we are revolving about the y-axis we use shells. When using shells, the cross sections are parallel to the axis about which we are

revolving. Therefore, we integrate w.r.t x. Your set up is incorrect. If solving \(\displaystyle y=x^{\frac{4}{9}}\) for y, we get \(\displaystyle x=y^{\frac{9}{4}}\).

\(\displaystyle 2{\pi}\int_{0}^{1}x\cdot x^{\frac{4}{9}}dx\)