Question

Find the area of the region{(x,y):y^2<6ax and x^2+y^2<16a^2} using the method of integration

Answer 1

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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