Calculus 2 Quiz: Area Between Parametric Curves

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Area Between Parametric CurvesQuestion 1 of 20

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:

\int_0^{\pi/2} a\sin^3 t \cdot (-3a\cos^2 t \sin t) \, dt

\int_0^{\pi/2} a\sin^3 t \cdot 3a\cos^2 t \sin t \, dt

\int_0^a y \, dx

where

y = a\left(1 - \left(\frac{x}{a}\right)^{2/3}\right)^{3/2}

\int_0^{\pi/2} |a\sin^3 t| \cdot |{-3a\cos^2 t \sin t}| \, dt

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Calculus 2 Quiz: Area Between Parametric Curves

Practice Area Between Parametric Curves in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Area Between Parametric Curves, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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All questions

Question 1

$\int_0^{\pi/2} a\sin^3 t \cdot (-3a\cos^2 t \sin t) \, dt$
$\int_0^{\pi/2} a\sin^3 t \cdot 3a\cos^2 t \sin t \, dt$
$\int_0^a y \, dx$ where $y = a\left(1 - \left(\frac{x}{a}\right)^{2/3}\right)^{3/2}$
$\int_0^{\pi/2} |a\sin^3 t| \cdot |{-3a\cos^2 t \sin t}| \, dt$ (correct answer)

Explanation: 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.

Question 2

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?

$\int_{-5}^{5} y \frac{dx}{dt} dt = \int_{-3}^{3} 2t \cdot 2t \, dt$
$\int_{-5}^{5} |y| \, dx = \int_{-3}^{3} |2t| \cdot 2t \, dt$
$\int_{-3}^{3} y \frac{dx}{dt} dt = \int_{-3}^{3} 2t \cdot 2t \, dt$
$\int_{-3}^{3} |y| \left|\frac{dx}{dt}\right| dt = \int_{-3}^{3} |2t| \cdot |2t| \, dt$ (correct answer)

Explanation: 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.

Question 3

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:

$\int_0^\pi 2\sin t \cdot (-3\sin t) \, dt = -6\int_0^\pi \sin^2 t \, dt$
$\int_0^\pi |2\sin t| \cdot |{-3\sin t}| \, dt = 6\int_0^\pi \sin^2 t \, dt$ (correct answer)
$\int_{-3}^3 2\sqrt{1 - \frac{x^2}{9}} \, dx$ after eliminating the parameter
$\int_0^\pi 2\sin t \cdot 3\cos t \, dt = 6\int_0^\pi \sin t \cos t \, dt$

Explanation: 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.

Question 4

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?

$\int_0^2 (t^2+1)e^t \, dt$
$\int_0^4 e^t(2t) \, dt$
$\int_1^5 e^{\sqrt{y-1}} \, dy$
$\int_0^2 e^t(2t) \, dt$ (correct answer)

Explanation: 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.

Question 5

Two curves, $C_1: x_1(t)=t, y_1(t)=2t$ and $C_2: x_2(t)=t^2, y_2(t)=t+1$ , enclose a region. They intersect at $t=1$ (at point (1,2)) and another point. To find the area, a student sets up $\int_{t_a}^{t_b} (y_1(t)x_1'(t) - y_2(t)x_2'(t)) \, dt$ . What must be true about the interval $[t_a, t_b]$ and the parameterizations for this setup to be potentially correct?

$[t_a, t_b]$ must be the interval in $t$ over which both curves trace the boundary of the region in the same direction with respect to $x$ .
The setup is fundamentally flawed because the curves have different parameterizations; the area must be calculated with a common variable like $x$ .
$[t_a, t_b]$ must correspond to the same $x$ -interval for both curves, and $C_1$ must be the upper curve while both are traced left-to-right. (correct answer)
The setup assumes that $x_1(t) = x_2(t)$ over the interval $[t_a, t_b]$ , which is not true for these specific curves.

Explanation: The integral

\int (y_{top} - y_{bottom}) \, dx

becomes

\int y_{top}(t)x'_{top}(t) dt - \int y_{bottom}(t)x'_{bottom}(t) dt

. For the student's combined integral

\int (y_1 x'_1 - y_2 x'_2) dt

to be correct, the limits of integration

[t_a, t_b]

must correspond to the same starting and ending

x

-values for both parameterizations. Furthermore, the subtraction

y_1 x'_1 - y_2 x'_2

implies that

C_1

is treated as the 'upper' curve and

C_2

as the 'lower' curve, and the standard subtraction format implies both are traced in the same direction (e.g., left-to-right) so that no sign change is needed to account for tracing direction.

Question 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 $\int (y_T(t)x_T'(t) - y_B(t)x_B'(t)) \, dt$ ?

When the region is symmetric with respect to the y-axis.
When the bounding curves can be more easily expressed as functions of $y$ rather than functions of $x$ . (correct answer)
When both curves are traced from left to right over the parameter interval.
When the region is located entirely in the first and fourth quadrants.

Explanation: 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

Question 7

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?

The curve must be traced in a clockwise direction.
The curve must be symmetric with respect to the origin.
The curve must be traced in a counter-clockwise direction. (correct answer)
The curve must lie entirely in the first quadrant.

Explanation: 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

\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.

Question 8

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?

$\int_0^1 t^2 \, dt$
$\int_0^1 \frac{\pi}{2} t^2 \cos(\frac{\pi}{2}t) \, dt$ (correct answer)
$\int_0^1 2t \sin(\frac{\pi}{2}t) \, dt$
$\int_0^1 \frac{\pi}{2} \cos(\frac{\pi}{2}t) \, dt$

Explanation: 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

Question 9

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?

The area is being calculated with respect to the y-axis.
The derivative $x'(t)$ is equal to 1 for the parameterization chosen.
The line $y=4$ is parameterized as $x_L(t)=t, y_L(t)=4$ .
Both B and C are implicit assumptions. (correct answer)

Explanation: 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

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.

Question 10

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?

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. (correct answer)
The curves intersect more than twice, so the region is not simple.
One curve should be parameterized with $t$ and the other with a different parameter like $s$ .
The formula $\int x \, dy$ cannot be used if $x$ is not a function of $y$ .

Explanation: 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.

Question 11

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?

The area must be calculated with $\int x \, dy$ for these functions.
The parameter $t$ represents different quantities in each parameterization, so they cannot be combined in a single integral with respect to $t$ . (correct answer)
The derivative $x'(t)$ is not constant for the cubic curve, which complicates the setup.
The intersection points in terms of $t$ are not the same for both curves, making the limits of integration ambiguous.

Explanation: 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.

Question 12

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?

The curves must have the same parameterization for their y-coordinates, i.e., $y_1(t)=y_2(t)$ .
The curves must have the same parameterization for their x-coordinates, i.e., $x_1(t)=x_2(t)$ . (correct answer)
The curves are traced in opposite directions over the interval $[a,b]$ .
The curves must be symmetric with respect to the x-axis.

Explanation: The formula for the area between two curves

y_{top}(x)

and

y_{bottom}(x)

\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

Question 13

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 $[\pi/2, 0]$ ?

They correspond to integrating with respect to $y$ from $y=0$ to $y=a$ .
They correspond to integrating with respect to $x$ from $x=0$ to $x=a$ . (correct answer)
They indicate that the curve is traced in a clockwise direction, which requires reversed limits.
They are an arbitrary choice; the integral would yield the same result if calculated from $0$ to $\pi/2$ .

Explanation: 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

Question 14

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?

The area must be calculated using $A = \int_0^{2\pi} x(t)y'(t) \, dt$ instead.
The result should be taken as its absolute value, since the formula gives signed area.
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. (correct answer)
The limits of integration must be reversed to $[2\pi, 0]$ to account for the orientation of the curve.

Explanation: 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.

Question 15

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?

$\int_0^{2\pi} (b \sin t)(-a \sin t) \, dt$ (correct answer)
$-\int_0^{2\pi} y(t)x'(t) \, dt$
$\int_0^{2\pi} x(t)y'(t) \, dt$
$\frac{1}{2} \int_0^{2\pi} (x(t)y'(t) - y(t)x'(t)) \, dt$

Explanation: 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.

Question 16

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?

$x(t) \ge 0$ and the curve is traced from left to right as $t$ increases from $a$ to $b$ .
$x(t) \ge 0$ and $y(t)$ is strictly increasing on $[a, b]$ . (correct answer)
$y(t) \ge 0$ and $x(t)$ is strictly increasing on $[a, b]$ .
The curve must be a simple closed curve traced counter-clockwise.

Explanation: 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

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.

Question 17

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 $C_2: (x_2(t), y_2(t))$ ?

$x_1(t)=x_2(t)=e^t$ , and $C_2$ is the top curve for $t \in [0,1]$ .
$y_1(t)=y_2(t)=e^t$ , and $C_2$ is the right curve for $t \in [0,1]$ .
$x_1'(t)=x_2'(t)=e^t$ , $C_2$ is the top curve, and the x-interval corresponds to $t \in [0,1]$ . (correct answer)
The parameterization must be $x(t)=t$ and $x'(t)=e^t$ , which is a contradiction.

Explanation: 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)

y_{top}(t)

y_1(t)

y_{bottom}(t)

, and

e^t

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.

Question 18

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?

Find intersection points, then integrate $\int y \, dx$ over the entire parameter interval $[-2, 2]$
Find where $\frac{dx}{dt} = 0$ , then integrate $\int |y| \left|\frac{dx}{dt}\right| dt$ between these parameter values
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 (correct answer)
Convert to Cartesian form and integrate $\int_{x_{min}}^{x_{max}} |y_{upper} - y_{lower}| \, dx$

Explanation: 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.

Question 19

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?

Determine which curve is above the other by comparing $y_1(1)$ and $y_2(1)$ at the midpoint
Express both curves in terms of the same parameter by substituting $s = 2 - t$
Find all intersection points by solving the system $t = 2-s$ and $t^2 - 1 = s^2$ (correct answer)
Convert both curves to Cartesian form $y = f(x)$ to eliminate parameters entirely

Explanation: 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.

Question 20

Two parametric curves intersect: Curve 1: $x = t$ , $y = t^2$ and Curve 2: $x = s$ , $y = 2s - 1$ . If they intersect when $t = 2$ and $s = 2$ , and we want the area between them from $x = 0$ to $x = 2$ , which expression correctly represents this area?

$\int_0^2 |(2s - 1) - t^2| \, dx$ where $s$ and $t$ are functions of $x$
$\int_0^2 |t^2 - (2t - 1)| \, dt$ since both curves use $x = t$ (correct answer)
$\int_0^2 |(t^2) - (2s - 1)| \frac{dx}{dt} \, dt$ with $s = t$
$\int_0^2 |t^2 - (2x - 1)| \, dx$ where $x$ ranges from 0 to 2

Explanation: Since both curves can be parameterized with the same parameter (both use their parameter as x-coordinate), we can write both in terms of t: y₁ = t² and y₂ = 2t - 1. The area between them from x = 0 to x = 2 is ∫₀² |t² - (2t - 1)| dt. Choice A incorrectly mixes parameters. Choice C unnecessarily includes dx/dt = 1. Choice D incorrectly treats the second curve as y = 2x - 1 instead of maintaining parametric form.