VITEEE 2016 Paper

If curve r = a sin 3 θ
To trace the curve, we consider the following table:

Thus there is a loop between θ = 0 & θ =

π

3

as r varies from r = 0 to r = 0.

Hence , the area of the loop lying in the positive quadrant =

1

2

‌

π 3

∫

0

r2dθ
=

1

2

‌

π 3

∫

0

sin2 Φ .

1

3

dΦ
[On putting, 3θ = Φ ⇒ dθ =

1

3

dΦ]
=

a2

6

‌

π 2

∫

0

sin2 Φ dΦ
=

a2

6

.

π 2

∫

0

1−cos2Φ

2

dΦ [Since cos 2θ = 1 - 2 sin2 θ]
=

a2

12

. [Φ+

sin2Φ

2

]0

π

2

=

a2

12

.[

π

2

+sinπ] =

πa2

24