Respuesta :
The correct question is:
Find the area of the surface generated by revolving about the x-axis, the portion of the astroid
[tex]x^\frac{3}{2} + y^\frac{2}{3} = 1[/tex]
Answer: The Surface Area is
[tex]\frac{6}{5}\pi[/tex]
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
[tex]y = \left(1 - x^\frac{2}{3}\right)^\frac{3}{2}[/tex]
Next, we differentiate y with respect to x
[tex]\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}}[/tex]
Thus,
[tex]\left(\frac{dy}{dx}\right)^2 = \frac{\left(1 - x\right)^\frac{2}{3}}{x^\frac{2}{3}}}[/tex]
and so
[tex]1 + \left(\frac{dy}{dx}\right) ^2 \\ \\ = 1 +\frac{(1 - x)^\frac{2}{3}}{x^ \frac{2}{3}} \\ \\=\frac{1}{x^\frac{2}{3}}[/tex]
Therefore, the Surface Area is given as:
[tex]\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. \\ \\[/tex]
If we let
[tex]u = 1-x^\frac{2}{3}[/tex]
then
[tex]du = -\frac{2}{3}x^{-\frac{1}{3}}dx,[/tex]
so we see that
[tex]= -\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[/tex]
