Area via Green's Theorem - Calculus 3

1. A region is bounded by the parametric curve $x = 3t - t^3$, $y = 3t^2$ for $t \in [0, \sqrt{3}]$ and the $x$-axis. Using Green's theorem to find the area, what integral represents the contribution from the parametric curve portion?
   Answer: $\int_0^{\sqrt{3}} (3t - t^3) \cdot 6t \, dt$ When using Green's theorem to find area, you're applying the formula $\text{Area} = \frac{1}{2}\oint_C (x \, dy - y \, dx)$, where the integral is taken around the boundary of the region in a counterclockwise direction. For the parametric curve portion, you need to convert this line integral to a parameter integral. Since $x = 3t - t^3$ and $y = 3t^2$, you find the differentials: $dx = (3 - 3t^2)dt$ and $dy = 6t \, dt$. Substituting into the Green's theorem formula, the contribution from the parametric curve becomes: $\frac{1}{2}\int_0^{\sqrt{3}} [(3t - t^3)(6t) - (3t^2)(3 - 3t^2)] \, dt$ The first term gives us $\int_0^{\sqrt{3}} (3t - t^3) \cdot 6t \, dt$, which matches answer choice C. Looking at the wrong answers: Choice A, $\int_0^{\sqrt{3}} 3t^2 \cdot (3 - 3t^2) \, dt$, represents the $-y \, dx$ term but with the wrong sign and missing the factor of $\frac{1}{2}$. Choice B incorrectly uses $dy = 3t^2$ instead of $dy = 6t \, dt$. Choice D also uses the wrong differential, treating $dy$ as if it were just $y$ rather than the actual derivative. Study tip: When applying Green's theorem with parametric curves, always compute $dx$ and $dy$ by differentiating the parametric equations with respect to the parameter. Don't confuse the functions $x(t)$ and $y(t)$ with their differentials $dx$ and $dy$.
2. The area enclosed by the curve $x = t^2 - 1$, $y = t^3 - t$ for $t \in [-\sqrt{2}, \sqrt{2}]$ is computed using Green's theorem. However, this parameterization creates a self-intersecting curve. At what point does the curve intersect itself?
   Answer: $(0, 0)$ For self-intersection, we need different parameter values giving the same point: $t_1^2 - 1 = t_2^2 - 1$ and $t_1^3 - t_1 = t_2^3 - t_2$ where $t_1 \neq t_2$. The first equation gives $t_1^2 = t_2^2$, so $t_2 = -t_1$. Substituting into the second: $t_1^3 - t_1 = (-t_1)^3 - (-t_1) = -t_1^3 + t_1$, which gives $2t_1^3 - 2t_1 = 0$, so $t_1 = 0, \pm 1$. For $t_1 = 1, t_2 = -1$: $x = 1-1 = 0$, $y = 1-1 = 0$. The other choices correspond to endpoints or non-intersection points.
3. The area of the region enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is computed using Green's theorem with the line integral $\oint_C x \, dy$. If the parameterization $x = a\cos t$, $y = b\sin t$ for $0 \leq t \leq 2\pi$ is used, what is the resulting area?
   Answer: $\pi ab$ Using Green's theorem, area equals $\oint_C x \, dy$. With $x = a\cos t$ and $y = b\sin t$, we have $dy = b\cos t \, dt$. So the integral becomes $\int_0^{2\pi} (a\cos t)(b\cos t) \, dt = ab\int_0^{2\pi} \cos^2 t \, dt = ab \cdot \pi = \pi ab$. Choice B forgets that $\int_0^{2\pi} \cos^2 t \, dt = \pi$, not $2\pi$. Choice C incorrectly divides by 2. Choice D doubles the wrong result.
4. A region $R$ has boundary consisting of two parts: the curve $y = x^2$ from $(0,0)$ to $(2,4)$, and the line segment from $(2,4)$ to $(0,0)$. To find the area using Green's theorem with $\oint_C x \, dy$, the integral must be split into two parts. What is the correct setup?
   Answer: $\int_0^2 x \cdot 2x \, dx + \int_4^0 (\frac{y}{2}) \cdot dy$ For the parabolic part $y = x^2$ from $(0,0)$ to $(2,4)$: $dy = 2x \, dx$, giving $\int_0^2 x \cdot 2x \, dx$. For the line segment from $(2,4)$ to $(0,0)$: $x = \frac{y}{2}$ and we integrate from $y = 4$ to $y = 0$, giving $\int_4^0 (\frac{y}{2}) \, dy$. Choice A incorrectly parameterizes the line segment. Choice C assumes $dy = 0$ on the line, which is wrong. Choice D incorrectly parameterizes the parabolic portion.
5. The area of a region bounded by $x = a(\theta - \sin \theta)$, $y = a(1 - \cos \theta)$ for $\theta \in [0, 2\pi]$ (a cycloid arch) is calculated using Green's theorem. What is the correct expression for this area?
   Answer: $\int_0^{2\pi} a(\theta - \sin \theta) \cdot a \sin \theta \, d\theta$ Using Green's theorem with $A = \oint_C x \, dy$, we have $x = a(\theta - \sin \theta)$ and $dy = a \sin \theta \, d\theta$. The area is $\int_0^{2\pi} a(\theta - \sin \theta) \cdot a \sin \theta \, d\theta$. Choice A incorrectly includes the $\frac{1}{2}$ factor, which is only needed when using $\frac{1}{2}\oint_C (x \, dy - y \, dx)$. Choices B and D use wrong expressions for the integrand. The cycloid forms a complete arch from $\theta = 0$ to $\theta = 2\pi$, and we integrate along the entire boundary including the base line.
6. Consider the region inside the circle $x^2 + y^2 = 4$ but outside the circle $x^2 + y^2 = 1$. To find its area using Green's theorem with $\oint_C x \, dy$, what is the correct integral setup?
   Answer: $\int_0^{2\pi} 2\cos \theta \cdot (-2\sin \theta) \, d\theta + \int_0^{2\pi} \cos \theta \cdot \sin \theta \, d\theta$ For the outer circle (radius 2), parameterized counterclockwise: $x = 2\cos \theta$, $dy = -2\sin \theta \, d\theta$, giving $\int_0^{2\pi} 2\cos \theta \cdot (-2\sin \theta) \, d\theta$. For the inner circle (radius 1), we need clockwise orientation to maintain the correct boundary orientation for the annular region: $x = \cos(-\theta) = \cos \theta$, $dy = \sin \theta \, d\theta$, giving $\int_0^{2\pi} \cos \theta \cdot \sin \theta \, d\theta$. The total is the sum (not difference) due to orientation. Choices A and B use wrong differentials. Choice C has wrong orientation for inner boundary.
7. A region $R$ is bounded by the curves $y = \sin x$ and $y = \cos x$ between their intersection points in $[0, 2\pi]$. To apply Green's theorem for area calculation, how many separate integrals are needed due to the multiple intersections?
   Answer: Three integrals, corresponding to intervals between consecutive intersection points When applying Green's theorem to find areas between intersecting curves, you need to identify how many distinct regions are created by the intersection points, since each region requires its own integral setup. First, find where $\sin x = \cos x$ in $[0, 2\pi]$. This occurs when $\tan x = 1$, giving intersection points at $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$. These two intersection points divide the interval $[0, 2\pi]$ into three separate intervals: $[0, \frac{\pi}{4}]$, $[\frac{\pi}{4}, \frac{5\pi}{4}]$, and $[\frac{5\pi}{4}, 2\pi]$. In each interval, the curves form a different enclosed region with different orientations. For $[0, \frac{\pi}{4}]$, $\cos x > \sin x$; for $[\frac{\pi}{4}, \frac{5\pi}{4}]$, $\sin x > \cos x$; and for $[\frac{5\pi}{4}, 2\pi]$, $\cos x > \sin x$ again. Each region needs its own line integral around its boundary to apply Green's theorem correctly. Answer A is wrong because it oversimplifies by grouping regions where the same function is larger, but Green's theorem requires separate integrals for each connected region. Answer B is incorrect because Green's theorem doesn't automatically handle multiple disconnected components—you must integrate around each component's boundary separately. Answer C is wrong because it assumes four intervals, but two intersection points in $[0, 2\pi]$ create only three intervals, not four. Remember: when using Green's theorem for area between curves, count the number of separate regions created by intersection points—each needs its own integral.
8. A triangular region has vertices at $(0,0)$, $(3,0)$, and $(1,2)$. Using Green's theorem to find the area with $\frac{1}{2}\oint_C (x \, dy - y \, dx)$, the boundary consists of three line segments. What is the contribution from the segment connecting $(3,0)$ to $(1,2)$?
   Answer: $\int_0^1 \left[(3-2t) \cdot 2 - (2t) \cdot (-2)\right] dt$ Parameterize the segment from $(3,0)$ to $(1,2)$ as $x = 3-2t$, $y = 2t$ for $t \in [0,1]$. Then $dx = -2 \, dt$ and $dy = 2 \, dt$. So $x \, dy - y \, dx = (3-2t)(2) - (2t)(-2) = 6-4t+4t = 6$, giving $\int_0^1 6 \, dt$. But this matches choice B when expanded: $(3-2t) \cdot 2 - (2t) \cdot (-2) = 6-4t+4t = 6$. Choice A uses wrong parameterization. Choice C has sign errors. Choice D has wrong derivative signs.

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