Find the Exact Area of Region Bounded by Parametric Curve $x=\sin t$, $y=\sin2t$ and the x-axis

Question

The curve C has parametric equations $x = \sin t$, $y = \sin 2t$, $0 \leqslant t \leqslant \frac{\pi}{2}$. The finite region R is bounded by the curve and the x-axis. Find the exact area of R. (6 marks)

Accepted Answer

$\frac{4}{3}$

Answer

Step 1: Recall the formula for the area under a parametric curve: $A = \int_{t_1}^{t_2} y \frac{dx}{dt} dt$, where $t_1$ and $t_2$ are the parametric bounds corresponding to the region bounded by the curve and x-axis. Step 2: Calculate $\frac{dx}{dt}$ from $x = \sin t$: $\frac{dx}{dt} = \cos t$. Step 3: Substitute $y = \sin2t$, $\frac{dx}{dt}=\cos t$, and the bounds $t=0$ to $t=\frac{\pi}{2}$ into the area formula: $A = \int_{0}^{\frac{\pi}{2}} \sin2t \cdot \cos t dt$. Step 4: Use the double-angle identity $\sin2t = 2\sin t \cos t$ to rewrite the integrand: $A = \int_{0}^{\frac{\pi}{2}} 2\sin t \cos^2 t dt$. Step 5: Use substitution to solve the integral: let $u = \cos t$, so $\frac{du}{dt} = -\sin t$, meaning $-du = \sin t dt$. Adjust the bounds: when $t=0$, $u=\cos0=1$; when $t=\frac{\pi}{2}$, $u=\cos\frac{\pi}{2}=0$. Step 6: Rewrite the integral in terms of u: $A = 2\int_{1}^{0} u^2 (-du) = 2\int_{0}^{1} u^2 du$. Step 7: Evaluate the definite integral: $2\left[\frac{u^3}{3}\right]_{0}^{1} = 2\left(\frac{1^3}{3} - \frac{0^3}{3}\right) = 2 \cdot \frac{1}{3} = \frac{4}{3}$.