$\displaystyle\iint_{S}6y^{2}\cos{x}dxdy$, where $S$ is the surface enclosed by the curves $y=\sin{x}$, the X-axis, and the line $x=\dfrac{\pi}{2}$
The first thing I don't know about solving this integral is what should be the lower limit of $x$? Should it be $0$ or something else? Why?
Then, how to solve this integral? I need step by step solution.
First draw the integral region and then to find the boundaries draw a line parallel to the $x-$axis, passing through the region and see where it intersect the region. For example in your example it will intersect the region in $x = \arcsin(y)$ and $x=\frac{\pi}{2}$. So this means that the lower bound is $\arcsin(y)$, while the upper one is $\frac{\pi}{2}$. For the other bounds, go one dimension lower, i.e. project the region onto the $y-$axis and see in which interval $y$ is. It shouldn't be hard to obtain that:
$$\displaystyle\iint_{S}6y^{2}\cos(x)dxdy = \displaystyle\int_0^{1}\int_{\arcsin(y)}^\frac{\pi}{2}6y^{2}\cos(x)dxdy$$