Question

The whole area surrounded by the curve with the equations $x = a \cos^3 t, y = b \sin^3 t$ is:

• A. $\frac{3}{4}\pi ab$
• B. $\frac{3}{8}\pi ab$
• C. $\frac{3}{8}\pi a^2b$
• D. $\frac{3}{8} ab$

Hint:

The Wallis formula is a massive time-saver for definite integrals of sine and cosine products from $0$ to $\pi/2$. Remember to multiply by $\pi/2$ only if both powers ($m$ and $n$) are even integers.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The given parametric equations $x = a \cos^3 t$ and $y = b \sin^3 t$ represent a scaled astroid. Because the curve is symmetric about both the x and y axes, we can calculate the area in the first quadrant (where $t$ varies from $0$ to $\pi/2$) and multiply by 4.
Step 2: Key Formula or Approach:
The area $A$ enclosed by a parametric curve is given by: \[ A = 4 \int_{0}^{a} y \, dx \] where we evaluate the integral in the first quadrant. Since $x = a \cos^3 t$, as $x$ goes from $0$ to $a$, the parameter $t$ goes from $\pi/2$ to $0$. Thus, $dx = -3a \cos^2 t \sin t \, dt$.
Step 3: Detailed Explanation:
Let's set up the integral for the whole area: \[ A = 4 \int_{\pi/2}^{0} (b \sin^3 t) (-3a \cos^2 t \sin t \, dt) \] Swapping the limits removes the negative sign: \[ A = 4 \int_{0}^{\pi/2} 3ab \sin^4 t \cos^2 t \, dt = 12ab \int_{0}^{\pi/2} \sin^4 t \cos^2 t \, dt \] We can evaluate this using the Wallis formula (or Beta function) for integrals of the form $\int_{0}^{\pi/2} \sin^m t \cos^n t \, dt$: \[ \int_{0}^{\pi/2} \sin^m t \cos^n t \, dt = \frac{[(m-1)(m-3)\dots 1][(n-1)(n-3)\dots 1]}{(m+n)(m+n-2)\dots 2} \times \frac{\pi}{2} \] Here, $m = 4$ (even) and $n = 2$ (even), so we multiply by $\pi/2$: \[ \int_{0}^{\pi/2} \sin^4 t \cos^2 t \, dt = \frac{(4-1)(4-3) \times (2-1)}{(4+2)(4+2-2)(4+2-4)} \times \frac{\pi}{2} \] \[ = \frac{(3 \times 1) \times (1)}{6 \times 4 \times 2} \times \frac{\pi}{2} = \frac{3}{48} \times \frac{\pi}{2} = \frac{1}{16} \times \frac{\pi}{2} = \frac{\pi}{32} \] Now, multiply this back into our area equation: \[ A = 12ab \left( \frac{\pi}{32} \right) = \frac{12}{32} \pi ab = \frac{3}{8} \pi ab \] Step 4: Final Answer:
The whole area surrounded by the curve is $\frac{3}{8}\pi ab$.