areacalculuscirclesintegrationrectangles
I am being asked to calculate the pink area that is overlapping between the semicircle and the rectangle. I was only given the radius of the circle (5), the equation of a circle $(x^2+y^2 = r^2).$ I have found the intersection points between the circle and the rectangle $(-4,3)$ and $(4,3)$ but I am lost afterwards, I no longer know what to do.
The problem is symmetric when the shapes are centered to each other. As such, I can see 4 cases of overlap between a circle of radius $R$ and a rectangle of half lengths $a$ and $b$.
The shaded area corresponding to one quadrant only is calculated as follows: Consider the cases where $b \leq a$ and vary the radius $R$ from small to large.
$$ \text{(Area)} = \begin{cases} \tfrac{\pi}{4} R^2 & R \leq b \\ \tfrac{1}{2} b \sqrt{R^2-b^2} + \tfrac{1}{2} R^2 \sin^{-1}\left( \tfrac{b}{R} \right) & b < R \leq a \\ \tfrac{1}{2} \left( a \sqrt{R^2-a^2} + b \sqrt{R^2-b^2} \right) + \tfrac{1}{4} R^2 \left( 2 \sin^{-1}\left( \tfrac{a}{R} \right) + 2 \sin^{-1} \left( \tfrac{b}{R} \right)- \pi \right) & a < R \leq \sqrt{a^2+b^2} \\ a b & R > \sqrt{a^2+b^2} \end{cases} $$
You can check that at the transition values of $R$ the area results match between adjacent cases. For example area (2) when $R=b$ is $\tfrac{1}{2} b^2 \sin^{-1} \left( \tfrac{b}{b} \right) + \tfrac{1}{2} b \sqrt{b^2-b^2} = \tfrac{\pi}{4} b^2$ which equals the area (1) when $R=b$.
Depending on the value of $\theta$ and $\phi$, the locations of Y, U, M, V, Y may vary. We only discuss the simplest case. (I think the same logic is applicable to other cases.)
When $\theta$ and $\phi$ are known, U and V are found (and assumed) to be located on the top edge of the rectangle.
0) Find the co-ordinates of X, U, M, V, Y.
1) Find $[\triangle PUM]$.
2) Find all the sides of $\triangle PMV$.
3) Find $\alpha$ by cosine law.
4) Find [sector PMY].
Best Answer
There are several ways you could approach this problem, but I think this would be the most straight-forward: