volume of $\displaystyle y=x^2$ and y=4 rotated about the x-axis
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Originally Posted by yoman360 volume of $\displaystyle y=x^2$ and y=4 rotated about the x-axis washers ... $\displaystyle R = 4$ $\displaystyle r = x^2$
Originally Posted by skeeter washers ... $\displaystyle R = 4$ $\displaystyle r = x^2$ ok this is what i did $\displaystyle int_0^2 16-x^4 dx$ integrate i get $\displaystyle \frac{128 \pi}{5}$ correct?
Originally Posted by yoman360 ok this is what i did $\displaystyle int_0^2 16-x^4 dx$ integrate i get $\displaystyle \frac{128 \pi}{5}$ correct? That's just half the volume $\displaystyle V=\int_{-2}^2\pi(16-x^4) dx=2\int_0^2\pi(16-x^4) dx$
Last edited by ione; Apr 21st 2010 at 07:02 AM.
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