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This was a modification of a previous question I asked, except now I'm saying how to solve the area inside $$x^2+y^2+\sin (4x) +\sin (4y) = 4 $$ The equation is a simple closed shape. Here's is the link of my previous question?

Finding the area of a implicit relation

I am a calculus I student, I learned about implicit derivatives, I was wondering how to find the area under implicit closed shapes.I then found Green's theorem, however; its mathematics is way ahead of me.

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    $\begingroup$ Green's Theorem is a theorem that relates a double integral to a single integral. To utilize the theorem on finding the area, first set up a double integral of $1$ over the bounded region D. From the theorem you can find that the area is equal to $\oint_{\partial D}x dy$, where $\partial D$ is the positively oriented boundary of the region D. $\endgroup$ – Derek 朕會功夫 Nov 14 '14 at 10:21
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    $\begingroup$ (The hard part is parameterizing the curve.) $\endgroup$ – Derek 朕會功夫 Nov 14 '14 at 10:41
  • $\begingroup$ All I can say to help is that it is reflective at y=x. I remember I used reflections on the x-axis to parametrize $x^2+y^2+sin(4x)=4$.However the one in my question is different. $\endgroup$ – Arbuja Nov 14 '14 at 17:53
  • $\begingroup$ What methods can be used for parametrizing? $\endgroup$ – Arbuja Nov 14 '14 at 17:56
  • $\begingroup$ There's a thing called "change of coordinates" (in a single integral, this is called "u-substitution"), which by changing the coordinates system (surprise surprise), you can transform the shape to another shape that would be easier to parameterize and integrate. (Though I don't think it works in this case. You can try to change it into something similar to polar coordinates.) $\endgroup$ – Derek 朕會功夫 Nov 14 '14 at 18:08
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If anyone can compute the actual area with computer programming, I would like to check my answer.

Since my graphing calculator was primitive I had to split the implicit shape into one-one functions. Then I used regression to approximate these segments.

I ended up getting an approximate area of 13.079062616. Just in case can someone check the answer.

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    $\begingroup$ I compute the area in polar coordinates using following command in maxima, $$\begin{array}{l} \text{R(t) := find_root(4-r^2-sin(4*r*cos(t))-sin(4*r*sin(t)),r,1,3);}\\ \text{quad_qag(R(t)^2/2,t,0,2*%pi,3);} \end{array} $$ I get a number about $12.71096745048912$ instead. maxima claims the accuracy is $5.77 \times 10^{-8}$. I have doubt this number is that accurate. However, the first few digits should be accurate enough and the area should be smaller than $13$. $\endgroup$ – achille hui Nov 18 '14 at 20:19

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