Answer :
Answer:
[tex]A_{c} =3\pi r^{2}[/tex]
Step-by-step explanation:
We can define the area under arch of the cycloid as:
[tex]A_{c} = \int_{a}^{b} ydx[/tex]
Let's evaluate this integral between 0 and 2π and put it in terms of dθ, using the chain rule.
[tex]A_{c} = \int_{0}^{2\pi} y\frac{dx}{d\theta}d\theta[/tex] (1)
Taking the derivative of x we have:
[tex]\frac{dx}{d\theta} = r(1 - cos(\theta))[/tex] (2)
Now, we can put (2) in (1).
[tex]A_{c}=\int_{0}^{2\pi} r(1 - cos(\theta))\cdot r(1 - cos(\theta))d\theta = \int_{0}^{2\pi} r^{2}(1 - cos(\theta))^{2}d\theta[/tex]
We can solve the quadratic equation to solve this integral:
[tex]A_{c} = \int_{0}^{2\pi} r^{2}(1 - cos(\theta))^{2}d\theta=r^{2}\int_{0}^{2\pi} (1-2cos(\theta)+cos^{2}(\theta))d\theta [/tex]
Now, we just need to take this integral by the sum rule. Let's recall we can use integration by part to solve cos²(θ)dθ.
[tex]A_{c} = r^{2}(\theta |_{0}^{2\pi} - 2sin(\theta)|_{0}^{2\pi} + 0.5\theta|_{0}^{2\pi} - 0.25sin(2\theta)|_{0}^{2\pi})[/tex]
Finally, the area is:
[tex]A_{c} = r^{2}(2\pi + \pi)=3\pi r^{2}[/tex]
Have a nice day!