find the area enclosed by the curve r²-4cosø
r2=4cosϕA=2×12∫−π2π24cosϕ dϕ=4sinϕ∣−π2π2=8\displaystyle r^2 = 4\cos{\phi}\\ \begin{aligned} A &= 2 \times \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 4\cos{\phi} \,\,\mathrm{d}\phi \\&= 4 \sin{\phi}\vert_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \\&= 8 \end{aligned}r2=4cosϕA=2×21∫−2π2π4cosϕdϕ=4sinϕ∣−2π2π=8
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