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Below is the exact question and answer from my textbook:
Find the area of the region enclosed between the two curves $C_{1}$ and $C_{2}$ where $C_{1}$ has the polar equation $r = \sin\theta$ and $C_{2}$ has the polar equation $r = \cos\theta$.

answer is
$\frac{\pi}{8} - \frac{1}{16}$

I spend some time figuring this out...
At first I need find intersection (ie: $\sin\theta = \cos\theta$) between this 2 equations but this
obviously didn't made any sense.
But then how would i find lower and upper limit [a,b]
using the formula for area = $\int_a^b \frac{1}{2}(\sin\theta-\cos\theta)^2 \,d\theta$

i assume the textbook is asking area for this: overlay of both equation

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    $\begingroup$ The answer is actually (pi/8)- (1/4) $\endgroup$ – user321137 Mar 8 '16 at 23:23
  • $\begingroup$ yes the area is $\frac{\pi}{8}-\frac{1}{4}$ $\endgroup$ – palio Mar 14 '16 at 18:48
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The sine and the cosine are equal when $\theta=\pi/4$. The two curves are actually circles with radii 1/2 and center $(0,1/2)$ and $(1/2,0)$ for the $\sin$ and $\cos$ respectively. You can thus find the area by computing the following integral

$$\int_0^{\pi/4} (\sin\theta)^2 d\theta$$

in which I multiplied by $2$ exploiting symmetry.

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  • $\begingroup$ The answer I get with Wolframalpha is not quite what you have though. $\endgroup$ – Raskolnikov Jul 17 '12 at 14:22
  • $\begingroup$ shouldn't the integrand be $(\sin\theta - \cos\theta)^2$ $\endgroup$ – kypronite Jul 17 '12 at 14:42
  • $\begingroup$ No, I use the polar integral $\frac{1}{2}\int r^2 d\theta$. And I only integrate over half of the surface, which means I only need the curve $r=\sin\theta$. I then double to recover the full surface area. $\endgroup$ – Raskolnikov Jul 17 '12 at 14:47
  • $\begingroup$ I just edited my post to include image of overlay of both polar equations.Can you confirm that's what actually your solution give?Sorry,I'm still learning multivariable calculus... $\endgroup$ – kypronite Jul 17 '12 at 15:05
  • $\begingroup$ Yes, that's what I compute. $\endgroup$ – Raskolnikov Jul 17 '12 at 15:06
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I just added this plot to show what is Raskolnikov trying to tell you. You can find the area. Note the integral he wrote above.

enter image description here

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  • $\begingroup$ i already uploaded polar plot same time as yours.I think i try to visualize polar coordinate as unit circle, hence the confusion in my part. $\endgroup$ – kypronite Jul 17 '12 at 15:12
  • $\begingroup$ @kypronite: Yes I saw that. I hope polar plot help you. :) $\endgroup$ – mrs Jul 17 '12 at 15:14
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For the curve $r=\sin(\theta)$ $$ \left.\begin{array}{} x=r\cos(\theta)=\sin(\theta)\cos(\theta)\\ y=r\sin(\theta)=\sin^2(\theta) \end{array}\right\}\quad x^2+\left(y-\tfrac12\right)^2=\tfrac14 $$ For the curve $r=\cos(\theta)$ $$ \left.\begin{array}{} x=r\cos(\theta)=\cos^2(\theta)\\ y=r\sin(\theta)=\sin(\theta)\cos(\theta) \end{array}\right\}\quad\left(x-\tfrac12\right)^2+y^2=\tfrac14 $$ So the area is the intersection of the two circles:

enter image description here

The area of the purple lune is $\frac14$ of the area of a radius $\frac12$ circle minus the area of a $\frac12,\frac12,\frac1{\sqrt2}$ right triangle, that is $$ \frac14\cdot\frac\pi4-\frac12\cdot\frac12\cdot\frac12 $$ Since the area of the green lune is the same, the area of the given intersection is $$ \frac\pi8-\frac14 $$

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@robjohn's calculation states "pi/4 x 1/4 - 1/2 x 1/2 x 1/2 = pi/8 -1/4" .

This is false, it equals pi/16 -1/8.

Here's a simpler argument that requires NO CALCULUS. Take @robjohn's diagram. Notice that the full square has area 1/2 x 1/2 = 1/4.

Then, notice that one quarter of one of these circles has area pi/16 (since pi/4 divided by 4 is pi/16).

Take the following difference: 1/4 - pi/16 = (4-pi)/16 call this D

Now, the area of interest is the total area of the square (1/4) minus TWO of the D areas, i.e. 1/4 -2D = 1/4 -2( (4-pi)/16 ) = pi/8 - 1/4

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  • $\begingroup$ Your answer has been deleted once, it will still be deleted every time you create the exact same one. Instead of copying it again, maybe you should try to understand why it's being deletd. $\endgroup$ – Paul Hudford Apr 30 at 6:39
  • $\begingroup$ Greetings Thomas. I'm actually not sure why it's being deleted. Is there an error in my argument? I have pointed out an error in the above comment and offered a solution to the problem that does not require calculus. I believe that being able to solve a problem in a multitude of ways makes mathematics more appreciable and enjoyable. Furthermore, some may understand this proof easier than other proofs. $\endgroup$ – FractalDust Apr 30 at 7:01
  • $\begingroup$ Hi, there is no mistakes in the above answer. Contrary to your claim, it never states that $\frac{\pi}{4} \cdot \frac{1}{4}- \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{\pi}{8}-\frac{1}{4}$. It claims that this is the purple area. The total solution is twice this calculation, hence $ \frac{\pi}{8}-\frac{1}{4}$. Then "your" solution is a copy paste of rogjohn solution. Finaly you shoud format your answer with MathJax to make it readable. please see basic help on mathjax notation $\endgroup$ – Paul Hudford Apr 30 at 7:11
  • $\begingroup$ I see! Thank you very much for your clarification. I look forward to your mathematical criticism in the future. Thanks for helping me improve my mathematics. $\endgroup$ – FractalDust Apr 30 at 7:14
  • $\begingroup$ No problem, and welcome again to MSE. $\endgroup$ – Paul Hudford Apr 30 at 7:15

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