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Mathematics
Calculus
Finding the Area of a Figure Given by an Equation
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[QUOTE="WMDhamnekar, post: 6782190, member: 696551"] Question:- Find the area of the figure given by the cartesian equation below: $$\frac{(x+y)^2}{16}+\frac{(x-y)^2}{9}=1$$ Solution given:- Let $x+y= 4\cos{\alpha},x-y=3\sin{\alpha}$ Then $x=\frac{4\cos{\alpha}+3\sin{\alpha}}{2}$ $\Rightarrow dx=\frac{3\cos{\alpha}-4\sin{\alpha}}{2}d\alpha$ $y=\frac{4\cos{\alpha}-3\sin{\alpha}}{2}$. So, $ydx=\left(3-\frac{25}{8}*\sin{2\alpha}\right)*d\alpha$ [B]What is this ydx? Hence the required area is [/B] $$\displaystyle\int_0^{2\pi}\left(3-\frac{25}{8}*\sin{2\alpha}\right)*d\alpha$$ [B]Why does the author select ydx as integrand for computation of area of ellipse? What is the logic behind that?[/B] $=6\pi=18.849$ If any member of Math help Board knows the explanations for my queries, may reply to this question. [/QUOTE]
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Mathematics
Calculus
Finding the Area of a Figure Given by an Equation
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