Volume of Revolution: Cylinders vs Washers

[Asphyxiated]

Homework Statement

Find the volume of the solid bounded by the curves $$y = x^{1/3}$$ and $$y = x$$ when rotated around y=1.

Homework Equations

Volume with washers:

$$V = \pi \int R(x)^{2}-r(x)^{2} dx$$

where R(x) and r(x) are functions of x defining the inner and outer radii of the washers

Volume with cylinders:

$$V = 2 \pi \int R(x)*h(x) dx$$

where R(x) is a function of x (or just x) to define the radius of the cylinder and h(x) is a function of x to define the height of the cylinder (could be two functions minus one another)

The Attempt at a Solution

I did this both with cylinders and washers and get the right answer with cylinder but am 1/30 units off of the correct answer with washers, can anyone tell me why?

First with cylinders:

$$2 \pi \int_{0}^{1} y * (y-y^{3}) dy = 2 \pi \int_{0}^{1} y^{2}-y^{4}dy$$

$$2 \pi \int_{0}^{1} y^{2}-y^{4}dy = 2 \pi ( \frac {y^{3}}{3}-\frac {y^{5}}{5})$$

evaluate at 1 and 0:

$$2 \pi ( \frac {1}{3} - \frac {1}{5}) = \frac {4 \pi} {15}$$

which is correct, now with washers:

$$\pi \int_{0}^{1} (1-x)^{2} - (1-x^{1/3})^{2} dx$$

$$\pi \int_{0}^{1} 1-2x+x^{2}- (1-2x^{1/3}+x^{2/3}) dx$$

$$\pi \int_{0}^{1} -2x +x^{2}+2x^{1/3}-x^{2/3} dx$$

$$\pi ( -x^{2} + \frac {x^{3}}{3} + \frac {3x^{4/3}}{2} - \frac {3x^{5/3}}{5})$$

evaluate at 1 and 0:

$$\pi (-1+\frac{1}{3}+\frac{3}{2}-\frac{3}{5})=\frac{7 \pi}{30}$$

Also maple confirms that these are the answers to these two integrals so I must be setting up the second one wrong but i have it all drawn out huge on my blackboard and can not see how you would say anything is different. It must be though! Please help me see where I went wrong here!

thanks!

 

[JThompson]

Are you sure the answer you got using cylinders is correct? Since you are rotating the region about y=1, the radii of your cylinders should be 1-y rather than y.

 

[Asphyxiated]

Yeah man, i am sure that's what it says although I asked my teacher today and you are right the shells are suppose to be 1-y as the radius and the correct answer is 7pi/30. She said that they must have screwed up when filling in the answer.

Thanks for pointing out that if the radius was 1-y then it comes out equal to the first integral with respect to x.