# 7329769 - 1 Curves r=2sin and r=2sin2 0 2 Area of the...

This preview shows pages 1–2. Sign up to view the full content.

Curves = r 2sinθ and = r 2sin2θ , 0 θ π2 Area of the region outside the first curve and inside the second curve: The point of intersection of both curves: = => = 2sinθ 2sin2θ θ π3 For given case, 0 θ π3 Required area is: = A 0π32sinθ2sin2θrdrdθ = - 420π3sin22θ sin2θdθ = - - - 20π31 cos4θ2 1 cos2θ2dθ = - 0π3cos2θ cos4θdθ = - sin2θ2 sin4θ40π3 = A 338 2. Convergence of series: = ∞ n 1 15n = + + +…+ Sn 15 110 115 15n And, + - = ( + )+ ( + )+…+ ( + )> ( + )+ ( + )+…+ ( + ) Sn p Sn 15 n 1 15 n 2 15 n p 15 n p 15 n p 15 n p Now, + = = p5n p 110 when p n i.e. - > S2n Sn 110 Thus for < ε 110 , we cannot have - < S2n Sn ε for all n N . Hence series is divergent . 3. Series: = ∞ + + n 1 n 12nn 2 = + + an n 12nn 2 Now, →∞ + + = limn n 12nn 2 1 which is not equal to 0. Hence, series is divergent. Sum = DNE 4. Series: = ∞ + n 1 4n 15n By ratio test, →∞ + < limn an 1an 1 for convergence. Now,

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 07/22/2011 for the course ME 3 taught by Professor Prof.ramachandran during the Spring '11 term at Indian Institute of Technology, Kharagpur.

### Page1 / 3

7329769 - 1 Curves r=2sin and r=2sin2 0 2 Area of the...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online