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Homework Statement
Find the volume of the solid bounded by the curves [tex] y = x^{1/3} [/tex] and [tex] y = x [/tex] when rotated around y=1.
Homework Equations
Volume with washers:
[tex] V = \pi \int R(x)^{2}-r(x)^{2} dx [/tex]
where R(x) and r(x) are functions of x defining the inner and outer radii of the washers
Volume with cylinders:
[tex] V = 2 \pi \int R(x)*h(x) dx [/tex]
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:
[tex] 2 \pi \int_{0}^{1} y * (y-y^{3}) dy = 2 \pi \int_{0}^{1} y^{2}-y^{4}dy [/tex]
[tex] 2 \pi \int_{0}^{1} y^{2}-y^{4}dy = 2 \pi ( \frac {y^{3}}{3}-\frac {y^{5}}{5}) [/tex]
evaluate at 1 and 0:
[tex] 2 \pi ( \frac {1}{3} - \frac {1}{5}) = \frac {4 \pi} {15} [/tex]
which is correct, now with washers:
[tex] \pi \int_{0}^{1} (1-x)^{2} - (1-x^{1/3})^{2} dx[/tex]
[tex] \pi \int_{0}^{1} 1-2x+x^{2}- (1-2x^{1/3}+x^{2/3}) dx [/tex]
[tex] \pi \int_{0}^{1} -2x +x^{2}+2x^{1/3}-x^{2/3} dx [/tex]
[tex] \pi ( -x^{2} + \frac {x^{3}}{3} + \frac {3x^{4/3}}{2} - \frac {3x^{5/3}}{5}) [/tex]
evaluate at 1 and 0:
[tex] \pi (-1+\frac{1}{3}+\frac{3}{2}-\frac{3}{5})=\frac{7 \pi}{30} [/tex]
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!
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