Answer to Question #335068 in Calculus for Johannes Steven

Question #335068

Find the area of the region that lies inside the circle r = 3sin"\\theta" and outside the cardioid r = 1+sin"\\theta"

1
Expert's answer
2022-05-01T04:20:43-0400


The cardioid intersects with the circle


"3\\sin\\theta=1+\\sin \\theta"

"\\sin \\theta=\\dfrac{1}{2}"

The cardioid intersects with the circle at "(\\dfrac{3}{2},\\dfrac{\\pi}{6}),(\\dfrac{3}{2},\\dfrac{5\\pi}{6})" and the pole.

The area of interest has been shaded above.

To find the area of a polar curve, we use


"A=\\dfrac{1}{2}\\displaystyle\\int_{\\pi\/6}^{5\\pi\/6}((3\\sin \\theta)^2-(1+\\sin \\theta)^2)d\\theta"

"=\\dfrac{1}{2}\\displaystyle\\int_{\\pi\/6}^{5\\pi\/6}(8\\sin^2 \\theta-2\\sin\\theta-1)d\\theta"

"=\\dfrac{1}{2}\\displaystyle\\int_{\\pi\/6}^{5\\pi\/6}(3-4\\cos2 \\theta-2\\sin\\theta)d\\theta"

"=\\dfrac{1}{2}[3\\theta-2\\sin2\\theta+2\\cos \\theta]\\begin{matrix}\n 5\\pi\/6\\\\\n \\pi\/6\n\\end{matrix}"

"=\\dfrac{1}{2}(\\dfrac{5\\pi}{2}+\\sqrt{3}-\\sqrt{3}-\\dfrac{\\pi}{2}+\\sqrt{3}-\\sqrt{3})"

"=\\pi"

"\\pi" square units.


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