The question is:
Find the area of the surface obtained by rotating the curve abotu the x-axis.
$x = \frac13(y^2 + 2)^{3/2}$
$1 \le y \le 2$
I do not understand how they went from $(y^2 + 1)^2$ to the result after the integral which reads $y(y^2 + 1)dy$. I know you have to plug in the original equation into the formula and the resulting answer $(y^2 + 1)^2$, but how does that equal to $y(y^2 + 1)$?
Thank you for your time.
Also, I found this question on a video on YouTube, and they explained R(y) = y. Why is that? How do you choose that value in the formula, like R(y) or R(x)? I don't get how you choose that value.
The surface integral in this case represents a sum of the surface areas of rings stacked along the $x$-direction and is given by
$$S=\int_1^2 2\pi y(y^2 + 1)dy$$
where $2\pi y$ is the circumference of the ring with radius $y$ considering that the surface revolves around the $x$ axis and $\sqrt{1+(x_y')^2}dy = (y^2 + 1)dy$ is the width of the ring.
Together, the integrand $2\pi y(y^2 + 1)dy$ represents the surface area of the ring of radius $y$ and within $dy$. Then, the total surface is its integration from 1 to 2 in $y$.