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^(3)/(2) + y^(2)/(3) = 1`

Answer: The Surface Area is

`(6)/(5)\pi`

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^(2)/(3)\right)^(3)/(2)`

Next, we differentiate y with respect to x

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

Thus,

`\left((dy)/(dx)\right)^2 = (\left(1 - x\right)^(2)/(3))/(x^(2)/(3))}`

and so

`1 + \left((dy)/(dx)\right) ^2 \\ \\ = 1 +((1 - x)^(2)/(3))/(x^ (2)/(3)) \\ \\=(1)/(x^(2)/(3))`

Therefore, the Surface Area is given as:

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

If we let

`u = 1-x^(2)/(3)`

then

`du = -(2)/(3)x^{-(1)/(3)}dx,`

so we see that

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