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Find the area A enclosed by the lemniscate with equation r^(2)=81 cos(2theta). Choose your limits of integration carefully. The lemniscate (Give your answer to the nearest whole number.)

Find the area A A enclosed by the lemniscate with equation r2=81cos⁡(2θ) r^{2}=81 \cos (2 \theta) . Choose your limits of integration carefully.\newlineThe lemniscate\newline(Give your answer to the nearest whole number.)

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Find indefinite integrals using the power rule (Level 2)right chevron icon

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Q. Find the area A A enclosed by the lemniscate with equation r2=81cos⁡(2θ) r^{2}=81 \cos (2 \theta) . Choose your limits of integration carefully.\newlineThe lemniscate\newline(Give your answer to the nearest whole number.)
  1. Understand the equation: Step 11: Understand the equation of the lemniscate.\newlineThe given equation is r2=81cos⁡(2θ)r^2 = 81 \cos(2\theta). This is a lemniscate, which is a figure-eight shaped curve. We need to find the area enclosed by one loop of the lemniscate.
  2. Set up the integral: Step 22: Set up the integral for the area.\newlineThe area AA of one loop of a lemniscate given by r2=a2cos⁡(2θ)r^2 = a^2 \cos(2\theta) is A=12∫r2dθA = \frac{1}{2} \int r^2 d\theta. Here, a2=81a^2 = 81, so r2=81cos⁡(2θ)r^2 = 81 \cos(2\theta). We integrate from θ=−π4\theta = -\frac{\pi}{4} to θ=π4\theta = \frac{\pi}{4} to cover one loop.
  3. Calculate the integral: Step 33: Calculate the integral.\newlineA=12∫−π4π481cos⁡(2θ)dθ.A = \frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 81 \cos(2\theta) d\theta.\newline =12×81×∫−π4π4cos⁡(2θ)dθ.= \frac{1}{2} \times 81 \times \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos(2\theta) d\theta.\newline =40.5×[sin⁡(2θ)2]−π4π4.= 40.5 \times \left[\frac{\sin(2\theta)}{2}\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}.\newline =40.5×[sin⁡(π2)−sin⁡(−π2)2].= 40.5 \times \left[\frac{\sin(\frac{\pi}{2}) - \sin(-\frac{\pi}{2})}{2}\right].\newline =40.5×[1−(−1)2].= 40.5 \times \left[\frac{1 - (-1)}{2}\right].\newline =40.5×(22).= 40.5 \times \left(\frac{2}{2}\right).\newline =40.5.= 40.5.

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