Question

The area bounded by the curve x = a cos³ t, y = a sin³ t is –

• A. $\frac { 3 \pi \mathrm { a } ^ { 2 } } { 8 }$
• B. $\frac { 3 \pi \mathrm { a } ^ { 2 } } { 16 }$
• C. $\frac { 3 \pi \mathrm { a } ^ { 2 } } { 32 }$
• D. 3pa²

Answer 1

Answer: (A)

$\frac { 3 \pi \mathrm { a } ^ { 2 } } { 8 }$

Answer 2

Eliminating t, we have

x^2/3 + y^2/3 = a^2/3 x = 0 Ž y = ± a

y = 0 Ž x = ± a Symmetric about both the axis.

Required area = 4 $\int _ { 0 } ^ { a } y d x$ = 4$\int _ { \pi / 2 } ^ { 0 } \mathrm { y } \cdot \frac { \mathrm { dx } } { \mathrm { dt } } \mathrm { dt }$ y = a sin³ t, x = a cos³ t Ž dx/dt = –3a cos² t sin t = 4 t . (–3a cos² t . sin t) dt = 12 a² $\int _ { 0 } ^ { \pi / 2 } \sin ^ { 4 }$t . cos² t dt =

= $\frac { 6 \mathrm { a } ^ { 2 } \times \frac { 3 } { 2 } \times \frac { 1 } { 2 } \times \sqrt { \pi } \cdot \frac { 1 } { 2 } \sqrt { \pi } } { 3 \times 2 }$

= $\frac { 3 } { 8 }$p a²