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Question: Area of a loop of the curve r = a sin 3q is -...

Area of a loop of the curve r = a sin 3q is -

A

πa26\frac { \pi \mathrm { a } ^ { 2 } } { 6 }

B

πa28\frac { \pi \mathrm { a } ^ { 2 } } { 8 }

C

πa212\frac { \pi \mathrm { a } ^ { 2 } } { 12 }

D

πa224\frac { \pi \mathrm { a } ^ { 2 } } { 24 }

Answer

πa212\frac { \pi \mathrm { a } ^ { 2 } } { 12 }

Explanation

Solution

Area of one loop

= 12∫0π/3a2sin⁡23θ dθ\frac { 1 } { 2 } \int _ { 0 } ^ { \pi / 3 } \mathrm { a } ^ { 2 } \sin ^ { 2 } 3 \theta \mathrm {~d} \theta

= 12⋅13a2∫0πsin⁡2tdt\frac { 1 } { 2 } \cdot \frac { 1 } { 3 } \mathrm { a } ^ { 2 } \int _ { 0 } ^ { \pi } \sin ^ { 2 } \mathrm { tdt }where t = 3q

= 13\frac { 1 } { 3 }a2 = 13\frac { 1 } { 3 }a2 ×12\frac { 1 } { 2 }×π2\frac { \pi } { 2 } = πa212\frac { \pi \mathrm { a } ^ { 2 } } { 12 }