Question

Find the surface area of the function x= ^3 square root of y between x=0 and x=1 when revolved around the axis

Answer 1

Solution:

$x=\sqrt[3]{y} \\\Rightarrow y=x^3=f(x)$

Between $x=0,x=1$ , revolved around $x$-axis.

Surface area$=2\pi\int_a^bf(x)\sqrt{1+[f'(x)]^2}dx$

$=2\pi\int_0^1 x^3\sqrt{1+[3x^2]^2}dx \\=2\pi\int_0^1 x^3\sqrt{1+9x^4}dx\ ...(i)$

Put $1+9x^4=t$

$\Rightarrow 36x^3=\frac{dt}{dx} \\\Rightarrow x^3 dx=\frac{dt}{36}$

When $x=0,t=1$

When $x=1,t=10$

So, (i) becomes,

Surface area$=\frac{2\pi}{36}\int_1^{10}\sqrt{t}dt$

$=\frac{\pi}{18}[\frac23t^{3/2}]_1^{10} \\=\frac{\pi}{27}[{10}^{3/2}-1] \\=3.563\ units^2$