find the area of the region lies inside the first curve and

find the area of the region lies inside the first curve and outside the second curve. r=2sintheta r=3sintheta

Solution

Let\'s find the points where the two curves meet : 3 sin ? = 2 - sin ? 4 sin ? = 2 sin ? = 1/2 ? = sin?¹ (1/2) + 2kp = p/6 + 2kp ; so ?1 = p/6 or ? = p - sin?¹ (1/2) + 2kp = 5p/6 + 2kp ; so ?2 = 5p/6 r1 = 3 sin ? r2 = 2 - sin ? The area is A = ?(? = p/6 to 5p/6) ?(r = r2 to r1) r dr d? = = (1/2) ?(? = p/6 to 5p/6) r1² - r2² d? = = (1/2) ?(? = p/6 to 5p/6) 9sin²? - 4 + 4sin? - sin²? d? = = 4?(? = p/6 , 5p/6) sin²? d? + 2?(? = p/6 , 5p/6) sin? d? - 2?(? = p/6 , 5p/6) d? = = 2(? - sin ? cos ?) - 2 cos ? - 2? | (? = p/6 , 5p/6) = = - 2 cos ? ( sin ? + 1) | (? = p/6 , 5p/6) = = - 2 [ cos 5p/6 (sin 5p/6 + 1) - cos p/6 ( sin p/6 + 1) ] = = 3v3