1. ## surface area

The given curve is rotated about the y-axis. Find the area of the resulting surface.
0x4

can someone help me figure this out. I know what the answer is i just dont know how to come to it. I found dy=(1/2)x-(1/2x)dx and then took dy/dx=2pi(x)*sqrt(dx^2+dy^2) and solved...any help? my way got me a completely wrong answer

2. Originally Posted by ahawk1
The given curve is rotated about the y-axis. Find the area of the resulting surface.
0x4

can someone help me figure this out. I know what the answer is i just dont know how to come to it. I found dy=(1/2)x-(1/2x)dx and then took dy/dx=2pi(x)*sqrt(dx^2+dy^2) and solved...any help? my way got me a completely wrong answer
$\displaystyle A = 2\pi \int_0^4 x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$

$\displaystyle \frac{dy}{dx} = \frac{x}{2} - \frac{1}{2x}$

$\displaystyle \left(\frac{dy}{dx}\right)^2 = \frac{x^2}{4} - \frac{1}{2} + \frac{1}{4x^2}$

$\displaystyle 1 + \left[\frac{x^2}{4} - \frac{1}{2} + \frac{1}{4x^2}\right] = \frac{x^2}{4} + \frac{1}{2} + \frac{1}{4x^2} = \left(\frac{x}{2} + \frac{1}{2x} \right)^2$

$\displaystyle A = 2\pi \int_0^4 x \sqrt{\left(\frac{x}{2} + \frac{1}{2x} \right)^2} \, dx$

$\displaystyle A = 2\pi \int_0^4 x \left(\frac{x}{2} + \frac{1}{2x} \right) \, dx = 2\pi \int_0^4 \frac{x^2}{2} + \frac{1}{2} \, dx = \frac{76\pi}{3}$