$x = 1 + 2y^2$, $1 \leq y \leq 2$
I sketched out a graph plotted some points to get a rough picture of the graph. So first things first, I think I need to find the x bounds, so I plug in y=1 and y=2 to find x.
$x = 1 + 2(1)^2 = \textbf{3}$ and
$x = 1 + 2(2)^2 = \textbf{5}$
Here is where I get confused setting up the integral.
I know that the surface area formula is $$S = \int 2\pi r ds$$ where $2\pi r$ represents the circumference of the revolution around the x-axis and $ds$ is the "thickness" of the same circle.
So I reason the formula to be setup like this
$$S = 2\pi \int_{3}^{5} x \frac{dx}{dy} =2\pi \int_{3}^{5} (1 + 2y^2)\sqrt{1 + (4y)^2}dy$$.
Is this correct? If so, this looks kind of difficult to integrate. Would integration by parts be my best choice here?
As the limits of 3 and 5 are for the value of x, it should not be used in your integral for dy. In fact the integral should be
$$S = 2\pi\int yds $$
$$ = \int_{3}^{5} 2\pi \sqrt{\frac{x - 1}{2}} \sqrt{1 + (\frac{dy}{dx})^2} dx$$
$$ = \int_{3}^{5} \frac{\pi}{2} \sqrt{8x - 7} dx$$
$$ = \frac{\pi}{24} (33^{\frac{3}{2}} - 17^{\frac{3}{2}}) = 4.978\pi$$