revolution of an arc

[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 $\displaystyle x$-axis.

 

[khansaheb]

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 $\displaystyle x$-axis.

show us your effort/s to solve this problem.

 

[logistic_guy]

Surface Area $\displaystyle = \int_{0}^{1}2\pi f(x)\sqrt{1 + [f'(x)]^2} \ dx$

$\displaystyle f(x) = y = x^3$
$\displaystyle f'(x) = 3x^2$

Surface Area $\displaystyle = \int_{0}^{1}2\pi x^3\sqrt{1 + 9x^4} \ dx = \frac{\pi}{18}\int_{1}^{10}\sqrt{u} \ du = \frac{\pi}{18}\frac{2}{3}u^{3/2}\bigg|_{1}^{10} = \frac{\pi}{18}\frac{2}{3}\left(10\sqrt{10} - 1\right) \approx 3.6$

 

[RalphD]

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[Steven G]

First makes post with no work shown and then posts the solution. This is in violation of the posting guidelines.
Reported this OP to the admin---again

 

[khansaheb]

Surface Area $\displaystyle = \int_{0}^{1}2\pi f(x)\sqrt{1 + [f'(x)]^2} \ dx$

$\displaystyle f(x) = y = x^3$
$\displaystyle f'(x) = 3x^2$

Surface Area $\displaystyle = \int_{0}^{1}2\pi x^3\sqrt{1 + 9x^4} \ dx = \frac{\pi}{18}\int_{1}^{10}\sqrt{u} \ du = \frac{\pi}{18}\frac{2}{3}u^{3/2}\bigg|_{1}^{10} = \frac{\pi}{18}\frac{2}{3}\left(10\sqrt{10} - 1\right) \approx 3.6$

This am old problem

 

[logistic_guy]

This am old problem

That am old problem too.