Question

Find the area of the surface generated by revolving about the​ x-axis the portion of the astroid x Superscript two thirds Baseline plus y Superscript two thirds Baseline equals 9 Superscript two thirds equal to one x(3/2)+y(2/3)=1)

Answer 1

The correct question is:

Find the area of the surface generated by revolving about the​ x-axis, the portion of the astroid

$x^\frac{3}{2} + y^\frac{2}{3} = 1$

Answer: The Surface Area is

$\frac{6}{5}\pi$

Step-by-step explanation:

First, we rewrite the expression in terms of x,  because we are revolving about the x-axis, we want to integrate in terms of x. Doing that, we have

$y = \left(1 - x^\frac{2}{3}\right)^\frac{3}{2}$

Next, we differentiate y with respect to x

$\frac{dy}{dx} = \frac{3}{2}\left(1 - x^\frac{2}{3}\right)^\frac{1}{2}\left(-\frac{2}{3}\right)x^\frac{-1}{3}}\\ \\= -\frac{\sqrt{\left(1 - x\right)^\frac{2}{3}}}{x^\frac{1}{3}}$

Thus,

$\left(\frac{dy}{dx}\right)^2 = \frac{\left(1 - x\right)^\frac{2}{3}}{x^\frac{2}{3}}}$

and so

$1 + \left(\frac{dy}{dx}\right) ^2 \\ \\ = 1 +\frac{(1 - x)^\frac{2}{3}}{x^ \frac{2}{3}} \\ \\=\frac{1}{x^\frac{2}{3}}$

Therefore, the Surface Area is given as:

$\int_{0}^{1}2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2}dx \\ \\= \int_{0}^{1}2\pi\left(1 - x^\frac{2}{3}\right)^\frac{3}{2}\sqrt{\frac{1}{x^\frac{2}{3}}}dx \\ \\=\int_{0}^{1}2\pi\left(1-x^\frac{2}{3}\right)^\frac{3}{2}x^\frac{-3}{2}dx.$

If we let

$u = 1-x^\frac{2}{3}$

then

$du = -\frac{2}{3}x^{-\frac{1}{3}}dx,$

so we see that

$= -\frac{3}{2}\int_{0}^{1}2\pi\left(1-x^\frac{2}{3}\right)^\frac{3}{2} - \frac{2}{3}x^{-\frac{1}{3}} dx \\ \\= -3\pi \int_{0}^{1}u^\frac{3}{2}du \\ \\= 3\pi\frac{2}{5}u^\frac{5}{2}\left \{ {{u=1} \atop {u=0}} \right. \\ \\= \frac{6}{5}\pi$