Verified Solution • 2025–26 Session moderate QJEEM22Ma2465: JEE Main 2022 Mathematics - Class 12 JEEM22Ma2465 Solution: If the area of the region is A, then is equal to __________. If the area of the region {(x,y):x23+y23≤1, x+y≥0, y≥0}\left\{ {(x,y):{x^{{2 \over 3}}} + {y^{{2 \over 3}}} \le 1,\,x + y \ge 0,\,y \ge 0} \right\}{(x,y):x32+y32≤1,x+y≥0,y≥0} is A, then 256Aπ{{256A} \over \pi }π256A is equal to __________. Step-by-Step Verified Solution ∴\therefore∴ Area of shaded region =∫−(12)320((1−x23)32+x)dx+∫01(1−x23)32dx= \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {\left( {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}} + x} \right)dx + \int\limits_0^1 {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}}dx} }=−(21)23∫0((1−x32)23+x)dx+0∫1(1−x32)23dx =∫−(12)320(1−x23)32dx+∫−(12)320xdx= \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {{{\left( {1 - {x^{{2 \over 3}}}} \right)}^{{3 \over 2}}}dx + \int\limits_{ - {{\left( {{1 \over 2}} \right)}^{{3 \over 2}}}}^0 {xdx} }=−(21)23∫0(1−x32)23dx+−(21)23∫0xdx Let x=sin3θx = {\sin ^3}\thetax=sin3θ ∴\therefore∴ dx=3sin2θcosθdθdx = 3{\sin ^2}\theta \cos \theta d\thetadx=3sin2θcosθdθ =∫−π4π23sin2θcos4θdθ+(0+116)= \int\limits_{ - {\pi \over 4}}^{{\pi \over 2}} {3{{\sin }^2}\theta {{\cos }^4}\theta d\theta + \left( {0 + {1 \over {16}}} \right)}=−4π∫2π3sin2θcos4θdθ+(0+161) =9π64+116−116=36π256=A= {{9\pi } \over {64}} + {1 \over {16}} - {1 \over {16}} = {{36\pi } \over {256}} = A=649π+161−161=25636π=A ∴\therefore∴ 256Aπ=36{{256A} \over \pi } = 36π256A=36 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring
