Exams Joint Entrance Examination Mathematics Area Under Curves Question 74 Question 74 of 230 Hard NUMERICAL
If the area bounded by the curve 2 y 2 = 3 x 2 y^{2}=3 x 2 y 2 = 3 x , lines x + y = 3 , y = 0 x+y=3, y=0 x + y = 3 , y = 0 and outside the circle ( x − 3 ) 2 + y 2 = 2 (x-3)^{2}+y^{2}=2 ( x − 3 ) 2 + y 2 = 2 is A \mathrm{A} A , then 4 ( π + 4 A ) 4(\pi+4 A) 4 ( π + 4 A ) is equal to \_\_\_\_\_\_\_\_\_\_\_\_.
Answer: 42
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y 2 = 3 x 2 , x + y = 3 , y = 0 2 y 2 = 3 ( 3 − y ) 2 y 2 + 3 y − 9 = 0 2 y 2 − 3 y + 6 y − 9 = 0 ( 2 y − 3 ) ( y + 2 ) = 0 ; y = 3 / 2 ( ∫ 3 2 0 ( x R − x 2 ) d y ) − A 1 = ∫ 3 2 0 ( ( 3 − y ) − 2 y 2 3 ) d y − π 8 ( 2 ) A = ( 3 y − y 2 2 − 2 y 3 9 ) 3 2 0 − π 4 4 A + π = 4 [ 9 2 − 9 8 − 3 4 ] = 21 2 = 10.50 ∴ 4 ( 4 A + π ) = 42 \begin{gathered} {y}^{2}=\frac{3x}{2},x+y=3,y=0 \\ 2{y}^{2}=3(3-y) \\ 2{y}^{2}+3y-9=0 \\ 2{y}^{2}-3y+6y-9=0 \\ (2y-3)(y+2)=0;y=3/2 \\ ({\int }_{\frac{3}{2}}^{0}({x}_{R}-{x}_{2})dy)-{A}_{1} \\ ={\int }_{\frac{3}{2}}^{0}((3-y)-\frac{2{y}^{2}}{3})dy-\frac{\pi }{8}(2) \\ A={(3y-\frac{{y}^{2}}{2}-\frac{2{y}^{3}}{9})}_{\frac{3}{2}}^{0}-\frac{\pi }{4} \\ 4A+\pi =4[\frac{9}{2}-\frac{9}{8}-\frac{3}{4}]=\frac{21}{2}=10.50 \\ \therefore 4(4A+\pi )=42 \end{gathered} y 2 = 2 3 x , x + y = 3 , y = 0 2 y 2 = 3 ( 3 − y ) 2 y 2 + 3 y − 9 = 0 2 y 2 − 3 y + 6 y − 9 = 0 ( 2 y − 3 ) ( y + 2 ) = 0 ; y = 3/2 ( ∫ 2 3 0 ( x R − x 2 ) d y ) − A 1 = ∫ 2 3 0 (( 3 − y ) − 3 2 y 2 ) d y − 8 π ( 2 ) A = ( 3 y − 2 y 2 − 9 2 y 3 ) 2 3 0 − 4 π 4 A + π = 4 [ 2 9 − 8 9 − 4 3 ] = 2 21 = 10.50 ∴ 4 ( 4 A + π ) = 42 Answer the question to unlock