Area Between Curves: Functions of y - AP Calculus AB

1. Set up the $dy$ integral for area between $x=\ln(y+2)$ (left) and $x=2$ (right) on $-1 \le y \le 2$.
   Answer: $\displaystyle \int_{-1}^{2}\big[2-\ln(y+2)\big]\,dy$ This question tests your ability to set up area integrals with respect to y, applying the fundamental right-minus-left subtraction principle. When computing area using dy integration, we must subtract the left boundary function from the right boundary function to obtain positive area. The problem identifies $x = \ln(y + 2)$ as the left curve and $x = 2$ as the right curve, so our integrand becomes $2 - \ln(y + 2)$. Choice A incorrectly uses $\ln(y + 2) - 2$, which reverses the subtraction and would yield negative area—a typical error when students forget to consider relative positions. To confirm the ordering, evaluate at $y = 0$: the left curve gives $x = \ln(2) \approx 0.693$ while the right curve gives $x = 2$, verifying that the constant function $x = 2$ is indeed to the right.
2. Which integral gives the area between $x=y^2-1$ (left) and $x=3y$ (right) for $0\le y\le 2$?
   Answer: $\displaystyle \int_{0}^{2}\big[3y-(y^2-1)\big]dy$ This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = 3y and the left curve is x = y² - 1, so the integrand is 3y - (y² - 1). The limits of integration are from y = 0 to y = 2, as specified. A tempting distractor is choice B, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.
3. Which integral represents the area between $x=y^2$ (right) and $x=y^2-4$ (left) for $-2\le y\le 2$?
   Answer: $\displaystyle \int_{-2}^{2}\big[y^2-(y^2-4)\big]dy$ This problem tests the skill of finding areas between curves by integrating with respect to y. The area is given by the integral from the lower to upper y-value of (x_right - x_left) dy. Here, x = $y^2$ is the right curve and x = $y^2$ - 4 is the left curve, so the integrand is $y^2$ - ($y^2$ - 4). The limits are from y=-2 to y=2, matching choice C. A tempting distractor is choice A, which subtracts left minus right, yielding a negative value instead of the positive area. Always evaluate the functions at a point within the interval to confirm which curve is on the right.
4. An expression for area between $x=y-4$ (left) and $x=y^2$ (right) on $1\le y\le 4$ is
   Answer: $\displaystyle \int_{1}^{4}\big[y^2-(y-4)\big]dy$ This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = y² and the left curve is x = y - 4, so the integrand is y² - (y - 4). The limits of integration are from y = 1 to y = 4, as specified. A tempting distractor is choice B, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.
5. Find the area setup with respect to $y$ between $x=\frac{1}{y+1}$ (right) and $x=y-1$ (left) for $0\le y\le 2$.
   Answer: $\displaystyle \int_{0}^{2}\big[\tfrac{1}{y+1}-(y-1)\big]\,dy$ This problem assesses your understanding of area integrals with respect to y, specifically the critical right-minus-left ordering. When finding area between curves using dy integration, we always compute (right function) - (left function) to ensure positive area values. Given that x = 1/(y + 1) is on the right and x = y - 1 is on the left, the correct integrand is 1/(y + 1) - (y - 1). Choice A reverses this to (y - 1) - 1/(y + 1), which would produce negative area—a common mistake when students write functions in the order they appear in the problem statement. To verify positioning, test at y = 1: the left curve gives x = 0 while the right curve gives x = 1/2, confirming that 1/(y + 1) is rightmost throughout the interval.
6. Find an expression for area between $x=\tan y$ (right) and $x=0$ (left) for $0\le y\le \frac{\pi}{4}$.
   Answer: $\displaystyle \int_{0}^{\pi/4}\big[\tan y-0\big]dy$ This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = tan y and the left curve is x = 0, so the integrand is tan y - 0. The limits of integration are from y = 0 to y = π/4, as specified. A tempting distractor is choice A, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.
7. An integral for the area between $x=\frac{1}{y+1}$ (right) and $x=0$ (left) on $1\le y\le 3$ is
   Answer: $\displaystyle \int_{1}^{3}\left[\frac{1}{y+1}-0\right]dy$ This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = 1/(y + 1) and the left curve is x = 0, so the integrand is 1/(y + 1) - 0. The limits of integration are from y = 1 to y = 3, as specified. A tempting distractor is choice A, which subtracts in the wrong order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.
8. Set up the area between $x=\ln(y+2)$ (left) and $x=y+1$ (right) for $0\le y\le 2$.
   Answer: $\displaystyle \int_{0}^{2}\big[(y+1)-\ln(y+2)\big]dy$ This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = y + 1 and the left curve is x = ln(y + 2), so the integrand is (y + 1) - ln(y + 2). The limits of integration are from y = 0 to y = 2, as specified. A tempting distractor is choice A, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.
9. Which integral gives the area between $x=2y+1$ (left) and $x=y^2+3$ (right) from $y=-1$ to $y=2$?
   Answer: $\displaystyle \int_{-1}^{2}\big[(y^2+3)-(2y+1)\big]\,dy$ This question requires setting up an area integral with respect to y using the fundamental principle of right minus left. When integrating along the y-axis, we subtract the left boundary function from the right boundary function to get positive area. The problem states that x = 2y + 1 is on the left and x = y² + 3 is on the right, meaning we need (y² + 3) - (2y + 1) as our integrand. Choice A incorrectly uses (2y + 1) - (y² + 3), which would yield negative area since it subtracts the rightmost function from the leftmost. To verify which function is truly on the right, pick a y-value like y = 0: the left curve gives x = 1 while the right curve gives x = 3, confirming our setup.
10. Find an area expression between $x=3-y^2$ (left) and $x=5$ (right) for $-1 \le y \le 1$.
    Answer: $\displaystyle \int_{-1}^{1}\big[5-(3-y^2)\big]dy$ This problem tests the skill of finding areas between curves by integrating with respect to y. The area is given by the integral from the lower to upper y-value of $(x_{\text{right}} - x_{\text{left}}) \, dy$. Here, $x = 5$ is the right curve and $x = 3 - y^2$ is the left curve, so the integrand is $5 - (3 - y^2)$. The limits are from $y=-1$ to $y=1$, matching choice B. A tempting distractor is choice A, which subtracts left minus right, yielding a negative value instead of the positive area. Always evaluate the functions at a point within the interval to confirm which curve is on the right.
11. Which integral represents the area between $x=y^2+3$ (right) and $x=2$ (left) for $0\le y\le 2$?
    Answer: $\displaystyle \int_{0}^{2}\big[(y^2+3)-2\big]dy$ This problem tests the skill of finding areas between curves by integrating with respect to y. The area is given by the integral from the lower to upper y-value of (x_right - x_left) dy. Here, x = $y^2$ + 3 is the right curve and x = 2 is the left curve, so the integrand is ($y^2$ + 3) - 2. The limits are from y=0 to y=2, matching choice C. A tempting distractor is choice A, which subtracts left minus right, yielding a negative value instead of the positive area. Always evaluate the functions at a point within the interval to confirm which curve is on the right.
12. Which integral gives area between $x=2y^2$ (left) and $x=8$ (right) for $-2\le y\le 2$?
    Answer: $\displaystyle \int_{-2}^{2}\big[8-(2y^2)\big]dy$ This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = 8 and the left curve is x = 2y², so the integrand is 8 - 2y². The limits of integration are from y = -2 to y = 2, as specified. A tempting distractor is choice A, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.
13. Which $dy$ integral represents the area between $x=y^2+2y$ (left) and $x=6-y$ (right) for $-1\le y\le 2$?
    Answer: $\displaystyle \int_{-1}^{2}\big[(6-y)-(y^2+2y)\big]\,dy$ This question assesses your understanding of area integrals with respect to y, emphasizing the critical right-minus-left subtraction order. When finding area between curves using dy integration, we must subtract the left boundary function from the right boundary function to ensure positive area values. The problem identifies x = y² + 2y as the left curve and x = 6 - y as the right curve, so our integrand becomes (6 - y) - (y² + 2y). Choice B incorrectly writes (y² + 2y) - (6 - y), reversing the subtraction and yielding negative area—a common mistake when students list functions in problem order rather than positional order. To verify positioning, test at y = 0: the left curve gives x = 0 while the right curve gives x = 6, confirming that 6 - y is indeed to the right throughout the interval.
14. Set up the area integral (in $y$) between $x=-y^2+4$ (right) and $x=y-2$ (left) for $0\le y\le 3$.
    Answer: $\displaystyle \int_{0}^{3}\big[(-y^2+4)-(y-2)\big]\,dy$ This problem tests your understanding of area integrals with respect to y, specifically the right-minus-left subtraction order. When finding area between curves using dy integration, we always compute (right function) - (left function) where right means larger x-values. The problem identifies x = -y² + 4 as the right curve and x = y - 2 as the left curve, so our integrand must be (-y² + 4) - (y - 2). Choice A reverses this order to (y - 2) - (-y² + 4), which would produce a negative result—a frequent error when students mechanically subtract without considering position. To remember the correct order, visualize or test: at y = 1, the left curve gives x = -1 while the right curve gives x = 3, confirming that -y² + 4 is indeed rightmost.
15. Find the area setup (with respect to $y$) between $x=y^2-1$ (left) and $x=3-y$ (right) for $-1 \le y \le 2$.
    Answer: $\displaystyle \int_{-1}^{2}\big[(3-y)-(y^2-1)\big]\,dy$ This problem tests your ability to set up area integrals with respect to y by identifying which function is on the right versus left. When integrating with respect to y, we use the formula (right function) - (left function), where "right" means larger x-values and "left" means smaller x-values. The problem explicitly tells us that $x = y^2 - 1$ is on the left and $x = 3 - y$ is on the right, so we need $(3 - y) - (y^2 - 1)$ in our integrand. Choice B reverses this subtraction, which would give a negative area—a common mistake when students forget the right-minus-left rule. To avoid confusion, always sketch the curves or evaluate at a test point: at $y = 0$, the left curve gives $x = -1$ and the right curve gives $x = 3$, confirming that $3 - y$ is indeed to the right.
16. Which integral represents area between $x=1-y$ (left) and $x=3+y^2$ (right) for $-2\le y\le 1$?
    Answer: $\displaystyle \int_{-2}^{1}\big[(3+y^2)-(1-y)\big]dy$ This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = 3 + y² and the left curve is x = 1 - y, so the integrand is (3 + y²) - (1 - y). The limits of integration are from y = -2 to y = 1, as specified. A tempting distractor is choice A, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.
17. Which integral gives the area between $x=3y+1$ (right) and $x=y^2$ (left) on $0\le y\le 2$?
    Answer: $\displaystyle \int_{0}^{2}\big[(3y+1)-y^2\big]dy$ This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = 3y + 1 and the left curve is x = y², so the integrand is (3y + 1) - y². The limits of integration are from y = 0 to y = 2, as specified. A tempting distractor is choice A, which reverses the subtraction and gives a negative value. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.
18. Find the correct $dy$ area expression between $x=\sqrt{y+1}$ (right) and $x=\frac{y}{2}$ (left) on $0\le y\le 3$.
    Answer: $\displaystyle \int_{0}^{3}\big[\sqrt{y+1}-\tfrac{y}{2}\big]\,dy$ This question assesses your ability to construct area integrals with respect to y using the right-minus-left principle. When integrating along the y-axis to find area, we must subtract the left boundary from the right boundary to ensure positive area values. Given that x = $\sqrt{y + 1}$ is on the right and x = $\frac{y}{2}$ is on the left, the correct integrand is $\sqrt{y + 1} - \frac{y}{2}$. Choice A incorrectly writes $\frac{y}{2} - \sqrt{y + 1}$, which reverses the subtraction and would yield negative area—a common mistake when students focus on the functions themselves rather than their relative positions. To verify positioning, evaluate at y = 3: the left function gives x = $\frac{3}{2} = 1.5$ while the right function gives x = $\sqrt{4} = 2$, confirming our setup.
19. Set up the area between $x=2\cos y$ (left) and $x=3$ (right) for $-\frac{\pi}{2}\le y\le \frac{\pi}{2}$.
    Answer: $\displaystyle \int_{-\pi/2}^{\pi/2}\big[3-2\cos y\big]dy$ This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is $x = 3$ and the left curve is $x = 2 \cos y$, so the integrand is $3 - 2 \cos y$. The limits of integration are from $y = -\frac{\pi}{2}$ to $y = \frac{\pi}{2}$, as specified. A tempting distractor is choice A, which reverses the order and gives negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.
20. Find an expression for the area bounded by $x=y^2-2y$ (left) and $x=4-y$ (right) for $-1\le y\le 2$.
    Answer: $\displaystyle \int_{-1}^{2}\big[(4-y)-(y^2-2y)\big]\,dy$ This problem requires using integration with respect to y to find the area between curves given as functions of y. The area is calculated by integrating the difference between the right curve and the left curve over the given y-interval. Here, the right curve is x = 4 - y and the left curve is x = y² - 2y, so the integrand is (4 - y) - (y² - 2y). The limits of integration are from y = -1 to y = 2, as specified. A tempting distractor is choice B, which reverses the subtraction order and yields a negative area. Always determine which curve is on the right by evaluating their x-values at a test point in the interval.

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