Specifically, what is the surface area of revolution of the function y = sin(x)+ cos(x) about the x-axis on the interval from $\frac{\pi}{4}$ and $\frac{5\pi}{4}$? The answer I got from Wolfram Alpha is approx. 3.5957. However, I am having trouble solving the integral itself after setting it up. I would greatly appreciate it if you could provide the steps so I can solve future problems of this type. Thanks.
EDIT: I think the integral should be $$2\pi\int_{\pi/4}^{5\pi/4}|(\sin x +\cos x )|\sqrt{2-\sin 2x }$$ but I am not entirely sure how to solve it.
Best Answer
Given $y(x) = \cos x+\sin x$, the surface integral is
$$S= 2\pi\int_{\pi/4}^{5\pi/4} |y|\sqrt{1+(y_x')^2}dx =4\pi\int_{\pi/4}^{3\pi/4} y\sqrt{1+(y_x')^2}dx$$
Let $t=y_x' = -\sin x + \cos x$, which leads to $dt = -(\cos x + \sin x)dx = -ydx$. Then, the integral becomes,
$$S = 4\pi\int_{0}^{\sqrt2} \sqrt{1+t^2}dt = 2\pi(\sqrt6 +\sinh^{-1}\sqrt2)$$
where the integral is solved with the substitution $t=\sinh u$.