Rotating a figure...

**bobsanchez** · May 19th 2010, 06:06 PM

Find the volume of f(x)= 2-x^2 and g(x)=1 when rotated around the line y= 1

Alright, so when I do this I get the answer 86(pi)/15. According to the teacher's answer, it should be 16(pi)/15. I can't figure out what I did wrong, so can someone please solve this out and let me know what I messed up? Thank you!

**skeeter** · May 19th 2010, 06:46 PM

Originally Posted by bobsanchez

Find the volume of f(x)= 2-x^2 and g(x)=1 when rotated around the line y= 1

Alright, so when I do this I get the answer 86(pi)/15. According to the teacher's answer, it should be 16(pi)/15. I can't figure out what I did wrong, so can someone please solve this out and let me know what I messed up? Thank you!

$V = \pi \int_{-1}^1 [(2-x^2)-1]^2 \, dx$

$V = 2\pi \int_0^1 (1-x^2)^2 \, dx$

$V = 2\pi \int_0^1 1 - 2x^2 + x^4 \, dx$

$V = 2\pi \left[x - \frac{2x^3}{3} + \frac{x^5}{5}\right]_0^1$

$V = 2\pi \left[1 - \frac{2}{3} + \frac{1}{5}\right] = 2\pi \cdot \frac{8}{15} = \frac{16\pi}{15}$

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