how to find overlapping areas of a circle and triangle

Question

I was asked in my calculus class to find the blue areas, two of which are overlapping with a circle and one of which is a part of a semi circle, I was given the equation of a circle $\ x^2+y^2 = r^2$, I was also told the radius $\ r = 5$, I have found the intercept points $\ (-4,3) , (4,3)$ After this I no longer know how to find the areas. I posted recently asking how to find the pink area, but I have now realized that I also need to find the blue areas and don't want to edit the previous question with the big change enter image description here

Answer 1

Consider first quarter of coordinate. Mark max. point of circle as E, it's intersection with x axis and line y=3 as C and A respectively. Also point B(4,0) and intersection of lines y=3 and x=5 as D.Now:

$f(x)=y=\sqrt{25-x^2}$

$g(x)=y=3$

$S_{ABC(pink)}=\int^5_4[f(x)=\sqrt{25-x^2 }]dx=\big[\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2} Sin^{-1}\frac{x}{5}\big]^5_4=2.1$

$\int^5_4 [g(x)= 3]= \big[3x\big]^5_4=3$

⇒ $S_{ADC(blue)}= 3-2.1=0.9$

$S_{EABO(pink)}=\int^4_0[g(x)=3]=\big[3x\big]^4_0=12$;

⇒ $S_{EACO(pink)}=12+0.9=12.9$

$S_{FAE(blue)}=S_{FCO}-S_{EACO}$

$S_{FCO}=\int^5_0\sqrt{25-x^2 }dx=\big[\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2} Sin^{-1}\frac{x}{5}\big]^5_0=\frac{25}{2}\times \frac{\pi}{2}=19.6$

⇒$S_{FAE(blue)}=19.6-12.9=6.7$ ⇒$S_{blue}=2(6.7+2.1)=17.6$

$S_{(pink)}=2\times 12.9=25.8$