Solving this problem

Today, in this blog post, we are going to solve a problem that popped up in my recommendations (from Mind Your Decisions). While Mind Your Decisions has a video on how to work this mathematics problem, we are going to solve it... but with a twist: we must use calculus.

Doing the problem:

The first shaded area is bounded above the curve ${\displaystyle y=0}$ and bounded below by the curves ${\displaystyle y=\frac{x}{2}}$ and ${\displaystyle \frac{x^{2}}{9}+y^{2}=1}$.

The first logical thing to do is find where the bounded below curves intersect. It can be shown that the bounded below curves intersect at ${\displaystyle x=-\frac{6}{\sqrt{13}},y=-\frac{3}{\sqrt{13}}}$ and at ${\displaystyle x=\frac{6}{\sqrt{13}},y=\frac{3}{\sqrt{13}}}$. However, since the shaded areas are in the first quadrant of the coordinate plane, we should eliminate the former solution. So the latter solution prevails, namely ${\displaystyle x=\frac{6}{\sqrt{13}},y=\frac{3}{\sqrt{13}}}$.

The next logical thing to do is to find the first shaded area. Since there are two curves that bound the region from below, we split the region into two pieces by the line ${\displaystyle x=\frac{6}{\sqrt{13}}}$. Denote the shaded area as ${\displaystyle A_3}$, the shaded area to the left of ${\displaystyle x=\frac{6}{\sqrt{13}}}$ as ${\displaystyle A_1}$, and the shaded area to the right of ${\displaystyle x=\frac{6}{\sqrt{13}}}$ as ${\displaystyle A_2}$. Then it is axiomatic to conclude that ${\displaystyle A_{1}+A_{2}=A_{3}}$.

We can use an integral to conclude that ${\displaystyle A_{1}=\int_{0}^{\frac{6}{\sqrt{13}}}\frac{x}{2}dx=\frac{9}{13}}$.

We then switch our attention to find ${\displaystyle y}$ in the following equation: ${\displaystyle \frac{x^{2}}{9}+y^{2}=1}$; this is important because if we can find ${\displaystyle y}$, we can just plug and chug it in to the integrand, along with the proper bounds, append a differential, and we have our answer!

Solving for 𝑦 in this equation ${\displaystyle \frac{x^{2}}{9}+y^{2}=1}$ gives us ${\displaystyle y=\pm\frac{\sqrt{9-x^{2}}}{3}}$. However, for the reasons stated earlier, we have to eliminate the negative solution. So the positive solution prevails, namely ${\displaystyle y=\frac{\sqrt{9-x^{2}}}{3}}$.

We can now do the actions that we could not do earlier to conclude that ${\displaystyle A_{2}=\int_{\frac{6}{\sqrt{13}}}^{3}\frac{\sqrt{9-x^{2}}}{3}dx=-\frac{9}{13}+\frac{3\pi}{4}-\frac{3}{2}\sin^{-1}\left(\frac{2}{\sqrt{13}}\right)\approx0.7819}$.

Therefore, ${\displaystyle A_{3}=A_{1}+A_{2}=\frac{9}{13}+-\frac{9}{13}+\frac{3\pi}{4}-\frac{3}{2}\sin^{-1}\left(\frac{2}{\sqrt{13}}\right)=\frac{3\pi}{4}-\frac{3}{2}\sin^{-1}\left(\frac{2}{\sqrt{13}}\right)}$. This is the first shaded area, and the area we sought after, because we will equate the result in the next shaded area.

The next shaded area is very simple to calculate. It is bounded above ${\displaystyle \frac{x^{2}}{9}+y^{2}=1}$ but below ${\displaystyle y=mx}$.

Once again, the logical thing to do is find where the bounded below curves intersect. It can be shown that the bounded below curves intersect at ${\displaystyle x=3\sqrt{\frac{1}{9m^{2}+1}},y=3m\sqrt{\frac{1}{9m^{2}+1}}}$.

We could calculate the integral directly: ${\displaystyle \int_{0}^{3\sqrt{\frac{1}{1+9m^{2}}}}\left(\frac{\sqrt{9-x^{2}}}{3}-mx\right)dx}$, but the CAS (Computer Algebra System) today is very lazy, so we will use a ### (where ###${\displaystyle \ge{100}}$) IQ move to force it give an answer:

First, we will let ${\displaystyle a=3\sqrt{\frac{1}{1+9m^{2}}}}$. The resulting integral can be done with the CAS to obtain: ${\displaystyle \int_{0}^{a}\left(\frac{\sqrt{9-x^{2}}}{3}-mx\right)dx=\frac{1}{6}\left(a\left(\sqrt{9-a^{2}}-3am\right)+9\sin^{-1}\left(\frac{a}{3}\right)\right)}$.

Then, we tuck it back in to the Right Hand Side (RHS) to obtain:

${\displaystyle \frac{1}{6}\left(\left(3\sqrt{\frac{1}{1+9m^{2}}}\right)\left(\sqrt{9-\left(3\sqrt{\frac{1}{1+9m^{2}}}\right)^{2}}-3\left(3\sqrt{\frac{1}{1+9m^{2}}}\right)m\right)+9\sin^{-1}\left(\frac{1}{3}\left(3\sqrt{\frac{1}{1+9m^{2}}}\right)\right)\right)}$

And finally, we have distilled down the geometric nature of the problem to reveal it is an algebra problem in disguise:

${\displaystyle \frac{1}{6}\left(\left(3\sqrt{\frac{1}{1+9m^{2}}}\right)\left(\sqrt{9-\left(3\sqrt{\frac{1}{1+9m^{2}}}\right)^{2}}-3\left(3\sqrt{\frac{1}{1+9m^{2}}}\right)m\right)+9\sin^{-1}\left(\frac{1}{3}\left(3\sqrt{\frac{1}{1+9m^{2}}}\right)\right)\right)=\frac{3\pi}{4}-\frac{3}{2}\sin^{-1}\left(\frac{2}{\sqrt{13}}\right)}$

However, some people like the Left Hand Side (LHS) much cleaner, so assuming that ${\displaystyle m>0}$ (By the way, a valid assumption because the reasons stated earlier), we obtain the cleaner expression:

${\displaystyle \frac{3}{2}\csc^{-1}\left(\sqrt{1+9m^{2}}\right)=\frac{3\pi}{4}-\frac{3}{2}\sin^{-1}\left(\frac{2}{\sqrt{13}}\right)}$

We will leave up to you that the answer is ${\displaystyle m=\frac{2}{9}}$.