Question Number 101551 by mhmd last updated on 03/Jul/20

Answered by bobhans last updated on 03/Jul/20

(Q2) The area = ∫_0 ^1  ∫_0 ^x  dy dx = ∫_0 ^1  (y)]_0 ^x   dx  = ∫_0 ^1  x dx =  [ (1/2)x^2  ]_0 ^1 = (1/2) ♥

Commented bymhmd last updated on 03/Jul/20

sir i want in polar

Answered by mr W last updated on 04/Jul/20

Q2(b)  1−r cos θ=r sin θ  ⇒r=(1/(sin θ+cos θ))  A=∫_0 ^(π/2) ∫_0 ^(1/(sin θ+cos θ)) ρdρdθ  =(1/2)∫_0 ^(π/2) (1/((sin θ+cos θ)^2 ))dθ  =(1/2)∫_0 ^(π/2) (1/(1+sin 2θ))dθ  =(1/4)∫_0 ^(π/2) (1/(1+sin 2θ))d(2θ)  =(1/4)∫_0 ^π (1/(1+sin t))dt  =(1/4)[(2/(1+tan (t/2)))]_π ^0   =(1/2)(1−0)  =(1/2)