1. ## Help: Volumes!!!!

A region is bounded by x=1, y=8 and the curve y=x^3. Find the volume of the solid of revolution formed when this region is rotated about:
a) the y-axis
b) the x-axis

THNX

2. ## Re: Help: Volumes!!!!

Originally Posted by gnomeo
A region is bounded by x=1, y=8 and the curve y=x^3. Find the volume of the solid of revolution formed when this region is rotated about:
a) the y-axis
b) the x-axis

THNX
\noindent Well x^2 = y^{\frac{2}{3}} and so a rotation about the y-axis means, \\ \begin{align*} \quad V &= \pi \int_1^8 y^{\frac{2}{3}}\mathrm{d}y - \pi (1)^2(7) \\ &= \pi \left[\frac{3}{5}y^{\frac{5}{3}}\right]^8_1 - 7\pi \\ &= \pi \left(\frac{3}{5}(8)^{\frac{5}{3}} - \frac{3}{5}(1)^{\frac{5}{3}}\right) - 7\pi \\ &= \frac{58\pi}{3} \text{ cubic units} \end{align*} \\ And y^2 = x^6 so a rotation about the x-axis means, \\ \begin{align*} \quad V &= \pi (8)^2(1) - \pi \int_1^2 x^6 \mathrm{d}x \\ &= 64\pi - \pi \left[\frac{1}{7}x^7\right]^2_1 \\ &= 64\pi - \pi \left(\frac{1}{7}(2)^7 - \frac{1}{7}(1)^7\right) \\ &= \frac{321\pi}{7} \text{ cubic units} \end{align*} \\ Here, we just subtracted the volumes of cylinders because the integrals accounted for unwanted volume.

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