Find the area of the region that is bounded by the given curve and lies in the s

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Explanation / Answer

3p/2 ? (1/2) * r^2 d? p/2 3p/2 ? (1/2) * sin^2(?) d? ========> knowing the identity, sin^2(?) = (1/2) * (1 - cos(2?) ) p/2 3p/2 ? (1/2) * (1/2) * (1 - cos(2?) ) d? p/2 3p/2 ? (1/4) * (1 - cos(2?) ) d? p/2 3p/2 ? (1/4) - (1/4) * cos(2?) d? p/2 . . . . . . . . . . . . . . . . . . . . . 3p/2 (1/4) * ? - (1/4) * (1/2) * sin(2?) ] . . . . . . . . . . . . . . . . . . . . . . p/2 . . . . . . . . . . . . . . . 3p/2 (1/4) * ? - (1/8) * sin(2?) ] . . . . . . . . . . . . . . . . p/2 (1/4) * (3p/2 - p/2) - (1/8) * [ sin(2 * 3p/2) - sin(2 * p/2) ] (1/4) * - (1/8) * [ sin(3p/2) - sin(p/2) ] (p/2) - (1/8) * [ ( -v(3)/2 ) - ( v(3)/2 ) ] (p/) - (1/8) * [ ( -2v(3)/2 ) ] (p/2) - (1/8) * -v2(3) (p/2) + (v(3)/8) ˜ 0.74

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