MHB Finding Surface Area from Equation in Terms of x=g(y)

[ineedhelpnow]

if I am finding the surface area, and I am given the equation in term of x=g(y) about the x-axis, do i have to solve for y or can i just integrate in terms of y? would i just use dx/dy instead?

 

[ineedhelpnow]

that's how i did the problem:

$x=\frac{1}{3} (y^2+2)^{3/2}$ $1\le y \le 3$ about the x axis
$x'= \frac{(y^2+2)^{3/2}}{3} = \frac{ \frac{3(y^2+2)^{1/2}}{2}}{3}dy(y^2+2 ) = \frac{(y^2+2)^{1/2}*2y}{2} = y \sqrt{y^2+2}$

$S=\int_{1}^{3} \ 2\pi y \sqrt{1+y^2(y^2+2)}dy=2\pi \int_{1}^{3} \ y\sqrt{1+y^4+2y^2}dy=2\pi\int_{1}^{3} \ y\sqrt{(y^2+1)^2}dy=2\pi \int_{1}^{3} \ y(y^2+1)dy=2\pi \int_{1}^{3} \ (y^3+y) dy=2\pi [\frac{y^4}{4}+\frac{y^2}{2}]_{1}^{3} = 48 \pi$

is it correct?

 

[evinda]

that's how i did the problem:

$x=\frac{1}{3} (y^2+2)^{3/2}$ $1\le y \le 3$ about the x axis
$x'= \frac{(y^2+2)^{3/2}}{3} = \frac{ \frac{3(y^2+2)^{1/2}}{2}}{3}dy(y^2+2 ) = \frac{(y^2+2)^{1/2}*2y}{2} = y \sqrt{y^2+2}$

$S=\int_{1}^{3} \ 2\pi y \sqrt{1+y^2(y^2+2)}dy=2\pi \int_{1}^{3} \ y\sqrt{1+y^4+2y^2}dy=2\pi\int_{1}^{3} \ y\sqrt{(y^2+1)^2}dy=2\pi \int_{1}^{3} \ y(y^2+1)dy=2\pi \int_{1}^{3} \ (y^3+y) dy=2\pi [\frac{y^4}{4}+\frac{y^2}{2}]_{1}^{3} = 48 \pi$

is it correct?

It is correct! (Yes)