Hello I'm trying to solve this equation but kinda stuck.
The problem is as follows;
The region in the first quadrant bounded above by the line y =2, below by the curve y=2sinx, $0\le x\le\pi/2$, and on the left by the y-axis, about the line y=2
Now I'm bit confused on how to approach the problem because it is bit complicated when I have to get the volume about the y=2 axis since the equation is stated as a function of x.
Could someone please help me out to solve this problem?
The top border of the region is $y=2$. The bottom border is $y = 2 \sin x$. So the radius of the disk (with thickness $dx$) will be $2 - 2 \sin x$. From there you add up all of the disks in the region by integrating over $x$ from $0$ to $\pi / 2$.
The volume is then
$$V = \int_0^{\pi/2}\pi(2-2 \sin x)^2 dx.$$
Are you able to take it from there?