Surface area of revolution; $y=0.25x^3$, $0\le x\le 2$, revolved around $x$-axis

Question

I'm trying to solve the surface area of revolution for the function $y=0.25x^3$ revolved around $x$-axis from $0\le x\le 2$.

I got the answer $14.253$ using $$2\pi \int_0^2 0.25x^3 \sqrt{1+(0.75x^2)^2}\,dx$$

And according to my college prof I'm wrong.

Can someone help me find my mistake?

Answer 1

Here's a quick shot at it. Leave out the square root and integrate.

a = π/2 * x^4 / 4 | from 0 to 2

= 16 π / 8

= 2π

Do it the way you started. This is why I like coming here. I find out what I let get rusty. No areas of revolution integrals for almost 40 years....

Ok, let's try this:

a = 2π∫ 0->2 (1/4)*x^3/4 * (1 + 9/16 x^4)^(1/2) dx

Let's try d/dx (1 + 9/16 x^4)^(3/2)

we get 3/2*9/4*x^3*(1 + 9/16 x^4)^(1/2), or

   (27/8)*x^3*(1 + 9/16 x^4)^(1/2), so

a = 2π * 2/27 * (1 + 9/16 x^4)^(1/2) | 0->2

= 4π/27 * (10^(3/2) - 1)

which seems to be reasonable since the answer is a little bigger than 4π.