InterviewSolution
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InterviewSolution
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Using integration find the area of the region `{(x,y):x^2+y^2=ax,x,y>=0}` |
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Answer» `x^2+y^2<=2ax` `x^2-2ax+y^2+a^2-a^2<=0` `(x-a)^2+y^2<=a^2` `y^2>=ax` `(x-a)^2+ax=a^2` `x^2+a^2-2ax+ax=a^2` `x(x-a)=0` `x=0,a` `y^2=ax` `y^2=a*a=a^2` `y=a` Area of region`=int_0^(2a)ydx` `int_0^asqrt(ax)dx+int_a^(2a)sqrt(2ax-x^2)dx` `sqrtaint_0^asqrtxdx+int_a^(2a)sqrt(a^2-(x-a))dx` `2/3a^2+[0+(a^2pi)/4-0-0]` `=2/3a^2+pi/4a^2` `=a^2(2/3+pi/4)`. |
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