I’m stuck on this question.
The area enclosed between the curves $$y=\sin x \quad \text{and} \quad y=\cos x $$ and within the lines $x=\pi/6 $ and $x=\pi/4$ is rotated about the x-axis. Find the volume of the solid of revolution formed.
I need some help with the steps to integrate and solve this.
The above is the second part of a question whose first part is simply to find the area of the region. I was able to solve that as follows:
Between $x=\pi/6$ and $x=\pi/4$ we have $\cos(x)>\sin(x)$. So the area is
$$\begin{align} \int_{\pi/6}^{\pi/4} (\cos(x)-\sin(x)) dx &=[\sin(x)+\cos(x)]|_{π/6}^{π/4} \\ &=\sqrt{2}-\tfrac12-\tfrac12\sqrt{3} \end{align}$$
This is what I found to be the enclosed area between these curves.
For the second part, you can apply the washer method to integrate the revolving volume around $x$ as follows,
$$V=\int_{\pi/6}^{\pi/4}\pi (\cos^2 x - \sin^2 x) dx$$
where the integrand represents the area of circular rings with outer radius $\cos x$ and inner radius $\sin x$. Carry out the integral to get
$$V=\pi\int_{\pi/6}^{\pi/4}\cos 2x dx= \frac{\pi}{4}(2-\sqrt{3})$$