Find the area of the region{(x,y):y^2<6ax and x^2+y^2<16a^2} using the method of integration
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y2=6ax;x2+y2=16a2y2=6ax;x2+y2=16a2
y=6ax−−−√y=6ax
y=16a2−x2−−−−−−−−√y=16a2−x2
=x2+6a2−16a2=0=x2+6a2−16a2=0
=> (x+8a)(x−2a)=0(x+8a)(x−2a)=0
Hence the point of intersection (2a,±2a3–√)(2a,±2a3)
A=2[∫02ay1dx+∫2a4ay2dx]A=2[∫02ay1dx+∫2a4ay2dx]
=2×[∫02a6ax−−−√dx+∫2a4a16a2−x2−−−−−−−−√dx]=2×[∫02a6axdx+∫2a4a16a2−x2dx]
=2×6–√a∫02ax1/2dx+2∫2a4a16a2−x2−−−−−−−−√dx=2×6a∫02ax1/2dx+2∫2a4a16a2−x2dx
=16a23–√+16a2.π2=16a23+16a2.π2−8a23–√−16a2π6−8a23−16a2π6
=16a23–√=16a23−8a23–√+16a2(π2−π6)−8a23+16a2(π2−π6)
=16a23–√=16a23−8a23–√+16a2(2π6)−8a23+16a2(2π6)
=16a23–√=16a23−8a23–√+16a2π3
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