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$$\frac{\int_{-a}^a (-\frac ba \sqrt{a^2-x^2} + b\;)\mathrm{d}x + \frac{\pi ab}2}2=ab$$ The above equation finds the area of a rectangle given A and B (both > 0). I wrote this myself as a challenge. Using desmos I verified that it was true (the desmos). I was wondering if someone could write a proof for that because I suck at proofs have tried and failed to write one. The main issue is dealing with the square root part of the equation. If anyone would write a proof with a description of the process I would be very thankful.

Note:

This is not for homework, I am just making challenges for myself to find area of objects in new ways. I also wrote one for finding the area of a square using integrals and message me on chat or something if you want to see it.

How I found it:

I found the above equation by making a inscribed ellipse in a rectangle. I then found the area under the bottom half of the ellipse if you translate it to be tangent to the x axis. I added the area of the ellipse divided by 2 and divided the sum by 2 (why I did this is fuzzy because I had to tweak the formula to get it to be correct I am planning on asking about why it is like this later).

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  • $\begingroup$ I didn't put the equation in the title because I felt it would clutter it up if it should be there tell me and I will place it $\endgroup$ – Christopher Dec 8 '17 at 19:09
  • $\begingroup$ It's very cool that you put together this experiment. For the proof, it's sufficient to show $$\int_{-a}^{a}\frac{b}{a}\sqrt{a^2-x^2}\,dx = \frac{\pi ab}{2}$$ Can you do that? $\endgroup$ – Matthew Leingang Dec 8 '17 at 19:20
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    $\begingroup$ @MatthewLeingang OP struggled precisely with this part (sqrt). I don't think that equation would help him a lot. $\endgroup$ – TheSimpliFire Dec 8 '17 at 19:32
  • $\begingroup$ @TheSimpliFire: In the integral I wrote, the integrand is nonnegative, so the integral is too. I wanted to trim the algebra down to the essential part of the calculation. I should have noticed the OP said he was having trouble with precisely that part. Even so, I think simplifying it down to that one sqrt at least isolates the confusion. $\endgroup$ – Matthew Leingang Dec 8 '17 at 20:31
  • $\begingroup$ $$\int_{-a}^{a}\frac{b}{a}\sqrt{a^2-x^2}\,dx = \frac{\pi ab}{2}$$ is true since the area of the ellipse is simply $\pi ab$ $\endgroup$ – gimusi Dec 8 '17 at 20:50
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It is true:

We want to show, on rearranging your equality, that $$\int_{-a}^a \left(-\frac ba \sqrt{a^2-x^2} + b\right)\, \mathrm{d}x=\left(2-\frac\pi2\right)ab$$ The LHS is equivalent to $$-\frac ba \int_{-a}^a\sqrt{a^2-x^2}\, \mathrm{d}x+\left[bx\right]_{-a}^a=-\frac ba \left[\frac{x\sqrt{a^2-x^2}}2+\frac{a^2}2\sin^{-1}\frac xa\right]_{-a}^a+2ab$$ (integrate by parts by writing $f(x)=\sqrt{a^2-x^2}$ and $g'(x)=1$) or $$-\frac ba\left[\frac{a^2}2\cdot \frac \pi2-\frac{a^2}2\cdot\left(-\frac\pi2\right)\right]+2ab=2ab-\frac ba\left(\frac{\pi a^2}{2}\right)=\left(2-\frac\pi2\right)ab$$ as required.

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To evaluate $\int_{-a}^a \frac{b}{a}\sqrt{a^2-x^2}\,dx$, I recommend a trigonometric substitution $x= a \sin \theta$. Then $dx = a \cos\theta\,d\theta$ and $\sqrt{a^2-x^2} = a \cos\theta$. When $x=-a$, $\sin\theta = -1$, so $\theta =-\frac{\pi}{2}$. When $x=a$, $\sin\theta = 1$, so $\theta = \frac{\pi}{2}$. So \begin{align*} \int_{-a}^a \frac{b}{a}\sqrt{a^2-x^2}\,dx &= \int_{-\pi/2}^{\pi/2} \frac{b}{a} (a \cos\theta)(a \cos\theta\,d\theta) = ab \int_{-\pi/2}^{\pi/2} \cos^2\theta\,d\theta \end{align*} For this integral, we use a trig identity: $$ ab \int_{-\pi/2}^{\pi/2} \cos^2\theta\,d\theta = \frac{ab}{2}\int_{-\pi/2}^{\pi/2} (1+\cos2\theta)\,d\theta $$ Over the interval $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, $\cos2\theta$ makes a full period. So the integral of this piece is zero. What's left is $\int_{-\pi/2}^{\pi/2} 1\,d\theta$, which is clearly $\pi$. So $$ \frac{ab}{2}\int_{-\pi/2}^{\pi/2} (1+\cos2\theta)\,d\theta = \frac{\pi ab}{2} $$ and we are done.

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