Questions: Use the Green's Theorem area formula, Area of R=1/2 ∮(C) dy-y dx, to find the area of the region, R, enclosed by the astroid, r(t)=(2 cos^3 t) i+(2 sin^3 t) j such that 0 ≤ t ≤ 2π. The area of R is □ (Type an exact answer, using π as needed.)

Solution Steps

To find the area of the region enclosed by the astroid using Green's Theorem, we need to evaluate the line integral \(\frac{1}{2} \oint_{C} (x \, dy - y \, dx)\). The parametric equations for the astroid are given by \(x(t) = 2 \cos^3 t\) and \(y(t) = 2 \sin^3 t\). We will compute \(dx\) and \(dy\) as derivatives of \(x(t)\) and \(y(t)\) with respect to \(t\), respectively. Then, substitute these into the integral and evaluate it over the interval \(0 \leq t \leq 2\pi\).

The astroid is defined by the parametric equations: \[ x(t) = 2 \cos^3(t), \quad y(t) = 2 \sin^3(t) \] for \(0 \leq t \leq 2\pi\).

We compute the derivatives \(dx\) and \(dy\): \[ dx = \frac{dx}{dt} = -6 \sin(t) \cos^2(t), \quad dy = \frac{dy}{dt} = 6 \sin^2(t) \cos(t) \]

Using Green's Theorem, we evaluate the integral: \[ \oint_{C} (x \, dy - y \, dx) = \int_{0}^{2\pi} \left(2 \cos^3(t) \cdot 6 \sin^2(t) \cos(t) - 2 \sin^3(t) \cdot (-6 \sin(t) \cos^2(t))\right) dt \] This simplifies to: \[ \int_{0}^{2\pi} (12 \cos^4(t) \sin^2(t) + 12 \sin^4(t) \cos^2(t)) dt \] The result of this integral is: \[ \int_{0}^{2\pi} 3\pi \, dt = 3\pi \]

According to Green's Theorem, the area \(A\) of the region \(R\) is given by: \[ A = \frac{1}{2} \oint_{C} (x \, dy - y \, dx) = \frac{1}{2} \cdot 3\pi = \frac{3\pi}{2} \]

Final Answer