The area bounded by the 2 parabolas is A(θ) = 1/2∫(r₂²- r₁²).dθ between limits θ = a,b...
<span>the limits are solution to 3cosθ = 1+cosθ the points of intersection of curves. </span>
<span>2cosθ = 1 => θ = ±π/3 </span>
<span>A(θ) = 1/2∫(r₂²- r₁²).dθ = 1/2∫(3cosθ)² - (1+cosθ)².dθ </span>
<span>= 1/2∫(3cosθ)².dθ - 1/2∫(1+cosθ)².dθ </span>
<span>= 9/8[2θ + sin(2θ)] - 1/8[6θ + 8sinθ +sin(2θ)] .. </span>
<span>.............where I have used ∫(cosθ)².dθ=1/4[2θ + sin(2θ)] </span>
<span>= 3θ/2 +sin(2θ) - sin(θ) </span>
<span>Area = A(π/3) - A(-π/3) </span>
<span>= 3π/6 + sin(2π/3) -sin(π/3) - (-3π/6) - sin(-2π/3) + sin(-π/3) </span>
<span>= π.</span>
<span>4p−8 = 4 (p-2)
answer is </span><span>d. 4 (p-2)</span>
Answer:
The temperature of the chemical solution after t hours is 
Step-by-step explanation:
Given that the initial temperature of the chemical solution, 
.
The rate of decreasing temperature, 
So, after t hours, the temperature of the chemical solution will decrease by 
.
So, the temperature of the chemical solution after t hours,
Hence, the temperature of the chemical solution after t hours is 
.
See picture for solution steps and answer.
