revolution of an arc - 2

[logistic_guy]

Find the surface area generated when the given arc is revolved about the given axis.

\(\displaystyle y = x^3, \ \ 0 \leq x \leq 1\), about the \(\displaystyle y\)-axis.

 

[skeeter]

[imath]\displaystyle SA = 2\pi \int_a^b f(x) \sqrt{1+[f'(x)]^2} \, dx[/imath]

 

[logistic_guy]

[imath]\displaystyle SA = 2\pi \int_a^b f(x) \sqrt{1+[f'(x)]^2} \, dx[/imath]

Thanks for passing by professor skeeter.

Your attack to solve the problem was almost correct. The circumference of the circle about the \(\displaystyle x\)-axis is \(\displaystyle 2\pi f(x)\) while about the \(\displaystyle y\)-axis is \(\displaystyle 2\pi f(y)\).

So, \(\displaystyle SA\) should be:

\(\displaystyle 2\pi\int_{0}^{1}f(y)\sqrt{1 + [f'(y)]^2} \ dy\)

where \(\displaystyle f(y) = x = \sqrt[3]{y}\) and \(\displaystyle f'(y) = \frac{1}{3y^{2/3}}\).

Then,

\(\displaystyle SA = 2\pi\int_{0}^{1}y^{1/3}\sqrt{1 + \frac{1}{9y^{4/3}}} \ dy \approx 5.9\)

Solving the integral by hand is so tedious, so I used a \(\displaystyle \text{CAS}\).

 

[skeeter]

I read x-axis by mistake. My apologies.