Area Between Parametric Curves - Calculus 2

1. To find the area of the region bounded by the parabola $y=x^2$ and the line $y=4$, a student parameterizes the parabola as $x(t)=t, y(t)=t^2$ for $t \in [-2,2]$. They propose the integral $\int_{-2}^2 (4 - t^2) \, dt$. What is the implicit assumption made in this setup?
   Answer: Both B and C are implicit assumptions. The standard formula for the area between curves is $\int_a^b (y_{top} - y_{bottom}) \, dx$. Here, $y_{top}=4$ and $y_{bottom}=x^2$. The limits are from $x=-2$ to $x=2$. The integral is $\int_{-2}^2 (4-x^2) \, dx$. To convert this to a parametric integral using $x=t$, we have $dx = x'(t)dt = 1 \cdot dt$. The expression becomes $\int_{-2}^2 (4-t^2) \, dt$. This setup implicitly assumes that the line $y=4$ is also considered in terms of this parameterization (i.e., $y_{top}=4$ for each $t$ from -2 to 2) and that $x'(t)=1$. Therefore, both B and C are assumptions that lead to the simplified integral.
2. A closed curve $C$ is parameterized by $(x(t), y(t))$ for $t \in [a,b]$. The area enclosed is given by $A = \frac{1}{2} \int_a^b (x(t)y'(t) - y(t)x'(t)) \, dt$. This formula is a direct consequence of Green's Theorem. What condition on the curve $C$ is necessary for this formula to yield a positive area?
   Answer: The curve must be traced in a counter-clockwise direction. Green's Theorem states that $\oint_C P \, dx + Q \, dy = \iint_R (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \, dA$. The area of region $R$ is $\iint_R 1 \, dA$. We can choose $P$ and $Q$ such that $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1$. One common choice is $P = -y/2$ and $Q = x/2$. The line integral becomes $\oint_C -\frac{y}{2} \, dx + \frac{x}{2} \, dy$. In parametric form, this is $\int_a^b (-\frac{y(t)}{2}x'(t) + \frac{x(t)}{2}y'(t)) \, dt = \frac{1}{2} \int_a^b (x(t)y'(t) - y(t)x'(t)) \, dt$. By convention, Green's Theorem yields a positive result for a simple closed curve that is traversed in the counter-clockwise (positive) direction.
3. The area of the region enclosed by the curve $x(t)=2\cos(t)-\cos(2t), y(t)=2\sin(t)-\sin(2t)$ for $t \in [0, 2\pi]$ is to be calculated. If one uses the formula $A = \int_0^{2\pi} y(t)x'(t) \, dt$, the result is negative. What is the correct interpretation or modification needed?
   Answer: The integral $\int_0^{2\pi} y(t)x'(t) \, dt$ should be negated to get the correct positive area, because the curve is traced counter-clockwise. The formula $A = \oint y \, dx$ (or $\int y(t)x'(t) \, dt$) gives the area of a closed region. By convention (related to Green's Theorem), this integral is positive if the curve is traced clockwise and negative if traced counter-clockwise. The given curve (a deltoid) is traced counter-clockwise for $t \in [0, 2\pi]$. Therefore, the integral $\int_0^{2\pi} y(t)x'(t) \, dt$ will yield the negative of the geometric area. To obtain the positive area, one must calculate $-\int_0^{2\pi} y(t)x'(t) \, dt$. Option A is also a valid formula that would give the positive area directly, but C correctly explains why the original formula gives a negative result and what modification is needed.
4. The area of a region is calculated using the parametric integral $A = \int_a^b x(t)y'(t) \, dt$. For this integral to represent the area between the curve and the $y$-axis, which set of conditions is sufficient?
   Answer: $x(t) \ge 0$ and $y(t)$ is strictly increasing on $[a, b]$. The integral $A = \int_a^b x(t)y'(t) \, dt$ is the parametric form of $A = \int_c^d x(y) \, dy$. For this to represent the area between the curve and the y-axis, we need two conditions. First, the curve must lie to the right of the y-axis, so $x(t) \ge 0$. Second, the integral must be over an increasing range of $y$ values. This means $y(t)$ must be increasing as $t$ goes from $a$ to $b$, which ensures $dy = y'(t)dt$ is positive and we are integrating in the 'upward' direction. Therefore, $x(t) \ge 0$ and $y(t)$ being strictly increasing are sufficient conditions.
5. A parametric curve is given by $x = 3\cos t$, $y = 2\sin t$ for $0 ≤ t ≤ \pi$. The area enclosed by this curve and the x-axis is:
   Answer: $\int_0^\pi |2\sin t| \cdot |{-3\sin t}| \, dt = 6\int_0^\pi \sin^2 t \, dt$ This is the upper half of an ellipse. For the area between the parametric curve and x-axis, we use ∫|y||dx/dt|dt. Here y = 2sin t and dx/dt = -3sin t. Since we want area (positive), we need |2sin t|·|-3sin t| = 6sin²t for t ∈ [0,π]. Choice A gives negative area. Choice C uses Cartesian form but is valid too, however the parametric approach in B is more direct. Choice D incorrectly uses dy/dt instead of dx/dt.
6. Consider setting up an integral for the area of a region bounded by two parametric curves, $C_1$ and $C_2$. When is it more advantageous to use the form $\int (x_R(t)y_R'(t) - x_L(t)y_L'(t)) \, dt$ over the form based on $\int (y_T(t)x_T'(t) - y_B(t)x_B'(t)) \, dt$?
   Answer: When the bounding curves can be more easily expressed as functions of $y$ rather than functions of $x$. The integral form $\int (x_R y'_R - x_L y'_L) dt$ is derived from integrating with respect to $y$: $\int (x_{right} - x_{left}) \, dy$. This approach is advantageous when the boundary of the region is simpler to describe with $x$ as a function of $y$. For example, if the curves do not pass the vertical line test but do pass the horizontal line test. The other integral form is derived from $\int (y_{top} - y_{bottom}) \, dx$, which is better when the curves are functions of $x$.
7. A parametric curve $C$ traces the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ counter-clockwise. A standard parameterization is $x(t) = a \cos t, y(t) = b \sin t$ for $t \in [0, 2\pi]$. Which integral setup for the area is INCORRECT?
   Answer: $\int_0^{2\pi} (b \sin t)(-a \sin t) \, dt$ For a counter-clockwise closed curve, the area can be calculated by $A = \oint x \, dy$, $A = -\oint y \, dx$, or $A = \frac{1}{2} \oint (x \, dy - y \, dx)$. Let's check each option. C: $\int_0^{2\pi} x(t)y'(t) \, dt = \int_0^{2\pi} (a \cos t)(b \cos t) \, dt = ab\int_0^{2\pi} \cos^2 t \, dt = ab\pi$. This is the correct area. D: This is the standard symmetric form of Green's theorem for area, which is correct for counter-clockwise traversal. B: $-\int_0^{2\pi} y(t)x'(t) \, dt = -\int_0^{2\pi} (b \sin t)(-a \sin t) \, dt = ab\int_0^{2\pi} \sin^2 t \, dt = ab\pi$. This is also correct. A: This is the integral $\int_0^{2\pi} y(t)x'(t) \, dt$. For a counter-clockwise curve, this integral $\oint y \, dx$ evaluates to the negative of the area. Therefore, this setup is incorrect for finding the (positive) area.
8. A parametric curve $x = e^t \cos t$, $y = e^t \sin t$ for $0 ≤ t ≤ 2\pi$ creates a spiral. The area between one complete revolution of this spiral and the x-axis requires which integral setup?
   Answer: $\int_0^{2\pi} e^t \sin t \cdot e^t(\cos t - \sin t) \, dt$ For area between a parametric curve and the x-axis, we use ∫y(dx/dt)dt. Here y = eᵗsin t and dx/dt = d/dt(eᵗcos t) = eᵗ(cos t - sin t). Since the spiral crosses the x-axis multiple times, we technically need to account for sign changes, but choice A gives the basic setup. Choice B incorrectly suggests we always need absolute values. Choice C has wrong approach and limits. Choice D uses eᵗcos t instead of dx/dt.
9. The area of the region in the first quadrant enclosed by the astroid $x(t) = a \cos^3 t, y(t) = a \sin^3 t$ is being set up. A valid integral for this area is $I = \int_{\pi/2}^0 y(t)x'(t) \, dt$. What is the geometric interpretation of the limits of integration $[\pi/2, 0]$?
   Answer: They correspond to integrating with respect to $x$ from $x=0$ to $x=a$. The integral is of the form $\int y \, dx$. The area in the first quadrant is bounded by the curve and the axes, which means we are calculating $\int_{x=0}^{x=a} y \, dx$. Let's check the parameter values. At $t=0$, $x(0)=a, y(0)=0$. At $t=\pi/2$, $x(\pi/2)=0, y(\pi/2)=a$. To integrate with respect to $x$ from 0 to $a$, we need to use the parameter values corresponding to these $x$ values. The parameter $t$ must go from $\pi/2$ (where $x=0$) to $0$ (where $x=a$). Thus, the limits $[\pi/2, 0]$ correspond to integrating with respect to $x$ from 0 to $a$.
10. A parametric curve is defined by $x = a\cos^3 t$, $y = a\sin^3 t$ for $0 ≤ t ≤ \frac{\pi}{2}$, where $a > 0$. This traces one quarter of an astroid. The area between this curve and the coordinate axes is:
    Answer: $\int_0^{\pi/2} |a\sin^3 t| \cdot |{-3a\cos^2 t \sin t}| \, dt$ For area between a parametric curve and the axes, we need ∫|y||dx/dt|dt. Here y = asin³t and dx/dt = d/dt(acos³t) = -3acos²t sin t. Since we want positive area, we need |asin³t|·|-3acos²t sin t| = 3a²sin⁴t cos²t. For t ∈ [0,π/2], sin t ≥ 0 and cos t ≥ 0, so the absolute values matter for the negative dx/dt. Choice A gives negative area. Choice B has wrong sign for dx/dt. Choice C is correct in Cartesian form but more complex.
11. A region R is bounded on the right by $C_R: x(t)=t, y(t)=t^3$ for $t \in [0, 2]$ and on the left by $C_L: x(t)=t, y(t)=4t$ for $t \in [0, 1]$. This description is problematic for setting up a single area integral. Why?
    Answer: The curves are not defined over the same parameter interval, making a single integral of the form $\int (x_R-x_L)y' dt$ impossible. To set up an area integral between two parametric curves, such as $\int (x_R(t)y'_R(t) - x_L(t)y'_L(t)) \, dt$, we typically need the parameter $t$ to sweep through a single, common interval that corresponds to the y-interval bounding the region. Here, the curves are defined over different $t$ intervals ($[0,2]$ and $[0,1]$). While one could re-parameterize the curves to have a common interval, the description as given is problematic because a single integral with a single parameter $t$ cannot be directly applied over a single interval.
12. A region in the first quadrant is bounded by $y=x^3$ and $y=x$. Using the parameterization $x(t)=t^2, y(t)=t^6$ for the cubic curve and $x(t)=t, y(t)=t$ for the line, a student attempts to set up the area integral. What is a primary difficulty in using these two different parameterizations directly?
    Answer: The parameter $t$ represents different quantities in each parameterization, so they cannot be combined in a single integral with respect to $t$. The area is $\int_0^1 (x-x^3) dx$. If we use $x_L(t)=t, y_L(t)=t$ for the line, the x-interval [0,1] corresponds to $t \in [0,1]$. If we use $x_C(t)=t^2, y_C(t)=t^6$ for the cubic, the x-interval [0,1] also corresponds to $t \in [0,1]$. The area could be set up as $\int_0^1 y_L(t)x_L'(t) dt - \int_0^1 y_C(t)x_C'(t) dt = \int_0^1 t(1) dt - \int_0^1 t^6(2t) dt$. However, the core conceptual problem is that the 't' in the first parameterization is simply $x$, while the 't' in the second is $\sqrt{x}$. They are not the same variable. A valid setup requires a consistent parameterization for the independent variable of integration. Combining them into one integral like $\int (y_L(t)x_L'(t) - y_C(t)x_C'(t)) dt$ is only valid if $t$ has a consistent meaning across both terms, which it doesn't here. So B and D are both relevant, but B is more fundamental.
13. The area of the region bounded by the curve $C: x(t) = e^t, y(t) = t^2+1$, the $y$-axis, and the lines $y=1$ and $y=5$ is to be calculated. Which of the following integrals correctly sets up this calculation?
    Answer: $\int_0^2 e^t(2t) \, dt$ The region is bounded by the y-axis ($x=0$) and the curve, between two horizontal lines. This suggests integrating with respect to $y$. The area formula is $A = \int_{y_1}^{y_2} x(y) \, dy$. Here, the $y$ limits are from 1 to 5. We need to express $x$ as a function of $y$, or use the parametric formula $A = \int_{t_1}^{t_2} x(t)y'(t) \, dt$. First, find the parameter limits. $y(t)=1 \implies t^2+1=1 \implies t=0$. $y(t)=5 \implies t^2+1=5 \implies t^2=4 \implies t=2$ (since $x=e^t > 0$, we assume $t \ge 0$). The derivative is $y'(t)=2t$. So the integral is $A = \int_0^2 x(t)y'(t) \, dt = \int_0^2 e^t(2t) \, dt$. Option C is also correct as $t = \sqrt{y-1}$, so $x(y) = e^{\sqrt{y-1}}$, but it's not a parametric setup. Options A and D are identical except for the explanation. Let me fix option B. Option D is the correct parametric setup.
14. Two parametric curves are given: $C_1: x = t, y = t^2 - 1$ and $C_2: x = 2-s, y = s^2$ for $0 ≤ t ≤ 2$ and $0 ≤ s ≤ 2$. To find the area between these curves, which step must be completed first?
    Answer: Find all intersection points by solving the system $t = 2-s$ and $t^2 - 1 = s^2$ When finding area between two parametric curves, we must first identify their intersection points to determine the proper limits of integration and regions where one curve is above the other. This requires solving the system of equations for when both x and y coordinates match. Choice A only checks one point. Choice B assumes a relationship that may not preserve the correct parameterization. Choice D eliminates useful parametric information.
15. A region's area is given by $\int_0^1 (y_2(t) - y_1(t)) e^t \, dt$. Given that this was derived from $\int (y_{top}-y_{bottom})dx$, what can be inferred about the parametric curves $C_1: (x_1(t), y_1(t))$ and $C_2: (x_2(t), y_2(t))$?
    Answer: $x_1'(t)=x_2'(t)=e^t$, $C_2$ is the top curve, and the x-interval corresponds to $t \in [0,1]$. The original integral form is $\int (y_{top} - y_{bottom}) \, dx$. In parametric terms, this becomes $\int (y_{top}(t) - y_{bottom}(t)) x'(t) \, dt$, assuming a common parameterization where $x$ is monotonic and traced in the same direction. Comparing $\int_0^1 (y_2(t) - y_1(t)) e^t \, dt$ with the general form, we can identify $y_2(t)$ as $y_{top}(t)$, $y_1(t)$ as $y_{bottom}(t)$, and $e^t$ as $x'(t)$. This means both curves must share a parameterization such that $x_1'(t) = x_2'(t) = e^t$, and the integration happens over a t-interval that corresponds to the desired x-interval.
16. Consider the parametric equations $x = t^3 - 3t$, $y = t^2$ for $-2 ≤ t ≤ 2$. This curve has a self-intersection. To find the area of the loop formed by this self-intersection, which approach is most appropriate?
    Answer: Find intersection points by solving $x(t_1) = x(t_2)$ and $y(t_1) = y(t_2)$, then integrate over the parameter interval corresponding to the loop For a self-intersecting parametric curve, we must first find the intersection points by solving x(t₁) = x(t₂) and y(t₁) = y(t₂) for different parameter values t₁ ≠ t₂. Then integrate |y|dx/dt|dt over the parameter interval that traces the loop once. Choice A integrates over the wrong interval. Choice B finds vertical tangents, not intersections. Choice D is difficult because the curve isn't a function of x.
17. To compute the area of a region bounded by $C_1: (x_1(t), y_1(t))$ and $C_2: (x_2(t), y_2(t))$, a valid setup is $\int_a^b (y_1(t)-y_2(t))x_1'(t) \, dt$. What does this setup imply about the two curves?
    Answer: The curves must have the same parameterization for their x-coordinates, i.e., $x_1(t)=x_2(t)$. The formula for the area between two curves $y_{top}(x)$ and $y_{bottom}(x)$ is $\int (y_{top} - y_{bottom}) dx$. If we parameterize using a common parameter $t$, this becomes $\int (y_{top}(t) - y_{bottom}(t)) x'(t) dt$. The student's setup is $\int_a^b (y_1(t)-y_2(t))x_1'(t) \, dt$. This implies that $C_1$ is the top curve, $C_2$ is the bottom curve, and they share a common x-parameterization such that $dx = x_1'(t) dt$ can be factored out. This requires $x_1(t) = x_2(t)$ (or at least $x_1'(t) = x_2'(t)$ and they trace the same x-interval), which means their x-coordinates must be the same for any given parameter value $t$.
18. The area of the region bounded by $x=0$, $x=1$, $y=0$, and the curve $x(t)=\sin(\frac{\pi}{2}t), y(t)=t^2$ is to be found. Which of the following definite integrals correctly represents this area?
    Answer: $\int_0^1 \frac{\pi}{2} t^2 \cos(\frac{\pi}{2}t) \, dt$ The area under a curve is given by $\int y \, dx$. In parametric form, this is $\int_{t_1}^{t_2} y(t)x'(t) \, dt$. First, determine the limits for the parameter $t$. The region is bounded by $x=0$ and $x=1$. When $x=0$, $\sin(\frac{\pi}{2}t)=0 \implies t=0$. When $x=1$, $\sin(\frac{\pi}{2}t)=1 \implies \frac{\pi}{2}t = \frac{\pi}{2} \implies t=1$. So the limits for $t$ are from 0 to 1. Next, find the derivative $x'(t)$: $x'(t) = \frac{d}{dt}(\sin(\frac{\pi}{2}t)) = \frac{\pi}{2}\cos(\frac{\pi}{2}t)$. Now, substitute into the formula: $A = \int_0^1 y(t)x'(t) \, dt = \int_0^1 t^2 \cdot \frac{\pi}{2}\cos(\frac{\pi}{2}t) \, dt$.
19. The curve $C$ is defined by $x(t) = 3t^2, y(t) = 2t^3$ for $t \ge 0$. Which integral represents the area of the region bounded by $C$, the x-axis, and the line $x=3$?
    Answer: $\int_0^1 (2t^3)(6t) \, dt$ The area under the curve is $\int y \, dx$. First, we need to find the limits of integration for the parameter $t$. The region starts at the origin $(0,0)$, which corresponds to $t=0$. The region ends at the line $x=3$. We find the corresponding $t$ value: $x(t) = 3t^2 = 3 \implies t^2 = 1 \implies t=1$ (since $t \ge 0$). The derivative is $x'(t) = 6t$. The area integral in parametric form is $\int_{t_1}^{t_2} y(t)x'(t) \, dt$. Substituting the functions and limits, we get $\int_0^1 (2t^3)(6t) \, dt$.
20. A parametric curve is given by $x = t^2 - 4$, $y = 2t$ for $-3 ≤ t ≤ 3$. To find the area between this curve and the x-axis, which integral setup is correct?
    Answer: $\int_{-3}^{3} |y| \left|\frac{dx}{dt}\right| dt = \int_{-3}^{3} |2t| \cdot |2t| \, dt$ For area between a parametric curve and the x-axis, we need $\int |y| \left|\frac{dx}{dt}\right| dt$. Here, $y = 2t$ and $\frac{dx}{dt} = 2t$, so the setup is $\int_{-3}^{3} |2t| \cdot |2t| \, dt$. We need absolute values for both y and dx/dt because area is always positive. Choice A omits absolute values and uses wrong limits. Choice B uses wrong limits for integration. Choice C omits necessary absolute values.

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