Question 32.2: The Area ofa Surface of Revolution Set up the integral for t......
The Area ofa Surface of Revolution
Set up the integral for the area of the surface formed by revolving the graph of y=x^{3},\ 0\leq x\leq1, about the x-axis.
In this case, the radius is r(x) = x³ . Since f′(x) = 3x² , we have
S\;=\;2\pi\int_{0}^{1}x^{3}\sqrt{1+\left(3x^{2}\right)^{2}}\;d x\;=\;2\pi\int_{0}^{1}x^{3}\sqrt{1+9x^{4}}\;d x\;.
This integral can be calculated using substitution (u = x^{4} ) , and the answer is \frac{\pi}{27}\biggl(10^{\frac{3}{2}}-1\biggr)\approx3.563\,.
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