Area Between Curves: Functions of y
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AP Calculus AB › Area Between Curves: Functions of y
An expression for area between $x=y-4$ (left) and $x=y^2$ (right) on $1\le y\le 4$ is
$\displaystyle \int_{1}^{4}\big[(y+4)-y^2\big]dy$
$\displaystyle \int_{1}^{4}\big[y^2-(y-4)\big]dy$
$\displaystyle \int_{1}^{4}\big[y^2-(y+4)\big]dy$
$\displaystyle \int_{1}^{4}\big[(y-4)-y^2\big]dy$
$\displaystyle \int_{4}^{1}\big[y^2-(y-4)\big]dy$
Explanation
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.
Which integral gives the area between $x=y^2-1$ (left) and $x=3y$ (right) for $0\le y\le 2$?
$\displaystyle \int_{0}^{2}\big[3y-(y^2-1)\big]dy$
$\displaystyle \int_{0}^{2}\big[3y-(y^2+1)\big]dy$
$\displaystyle \int_{0}^{2}\big[(3y-1)-(y^2-1)\big]dy$
$\displaystyle \int_{0}^{2}\big[(y^2-1)-3y\big]dy$
$\displaystyle \int_{2}^{0}\big[3y-(y^2-1)\big]dy$
Explanation
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.
Which integral represents the area between $x=\sqrt{y}$ (left) and $x=\frac{y}{2}+1$ (right) for $0\le y\le 4$?
$\displaystyle \int_{0}^{4}\left[\sqrt{y}-\left(\frac{y}{2}+1\right)\right]dy$
$\displaystyle \int_{0}^{4}\left[\left(\frac{y}{2}+1\right)-\sqrt{y}\right]dy$
$\displaystyle \int_{4}^{0}\left[\left(\frac{y}{2}+1\right)-\sqrt{y}\right]dy$
$\displaystyle \int_{0}^{4}\left[\left(\frac{y}{2}+1\right)-\sqrt{y+1}\right]dy$
$\displaystyle \int_{0}^{4}\left[\left(\frac{y}{2}-1\right)-\sqrt{y}\right]dy$
Explanation
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/2 + 1 and the left curve is x = √y, so the integrand is (y/2 + 1) - √y. The limits of integration are from y = 0 to y = 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.
Which integral gives area between $x=2y^2$ (left) and $x=8$ (right) for $-2\le y\le 2$?
$\displaystyle \int_{-2}^{2}\big[8-(2y^2)\big]dy$
$\displaystyle \int_{-2}^{2}\big[8-(y^2)\big]dy$
$\displaystyle \int_{2}^{-2}\big[8-(2y^2)\big]dy$
$\displaystyle \int_{-2}^{2}\big[8-(2y)\big]dy$
$\displaystyle \int_{-2}^{2}\big[(2y^2)-8\big]dy$
Explanation
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.
Set up the area between $x=\ln(y+1)+2$ (right) and $x=1$ (left) for $0\le y\le 3$.
$\displaystyle \int_{0}^{3}\big[(\ln(y-1)+2)-1\big]dy$
$\displaystyle \int_{3}^{0}\big[(\ln(y+1)+2)-1\big]dy$
$\displaystyle \int_{0}^{3}\big[1-(\ln(y+1)+2)\big]dy$
$\displaystyle \int_{0}^{3}\big[(\ln(y+1)-2)-1\big]dy$
$\displaystyle \int_{0}^{3}\big[(\ln(y+1)+2)-1\big]dy$
Explanation
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 = ln(y+1) + 2 is the right curve and x = 1 is the left curve, so the integrand is (ln(y+1) + 2) - 1. The limits are from y=0 to y=3, 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.
Which integral gives the area between $x=\frac{y+1}{2}$ (right) and $x=y-2$ (left) for $-1\le y\le 3$?
$\displaystyle \int_{-1}^{3}\left[(y-2)-\frac{y+1}{2}\right]dy$
$\displaystyle \int_{3}^{-1}\left[\frac{y+1}{2}-(y-2)\right]dy$
$\displaystyle \int_{-1}^{3}\left[\frac{y-1}{2}-(y-2)\right]dy$
$\displaystyle \int_{-1}^{3}\left[\frac{y+1}{2}-(y+2)\right]dy$
$\displaystyle \int_{-1}^{3}\left[\frac{y+1}{2}-(y-2)\right]dy$
Explanation
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+1)/2 is the right curve and x = y-2 is the left curve, so the integrand is (y+1)/2 - (y-2). The limits are from y=-1 to y=3, matching choice A. A tempting distractor is choice B, 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.
Which integral gives the area between $x=2y+1$ (left) and $x=y^2+3$ (right) from $y=-1$ to $y=2$?
$\displaystyle \int_{-2}^{1}\big[(y^2+3)-(2y+1)\big],dy$
$\displaystyle \int_{-1}^{2}\big[(2y+1)-(y^2+3)\big],dy$
$\displaystyle \int_{-1}^{2}\big[(y^2+3)-(2y+1)\big],dy$
$\displaystyle \int_{-1}^{2}\big[(y^2-3)-(2y+1)\big],dy$
$\displaystyle \int_{-1}^{2}\big[(y^2+3)-(2y-1)\big],dy$
Explanation
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.
Set up the $dy$ integral for area between $x=\ln(y+2)$ (left) and $x=2$ (right) on $-1 \le y \le 2$.
$$\displaystyle \int_{-1}^{2}\big[2-\ln(y+2)\big],dy$$
$$\displaystyle \int_{-1}^{2}\big[2-\ln(y-2)\big],dy$$
$$\displaystyle \int_{-2}^{1}\big[2-\ln(y+2)\big],dy$$
$$\displaystyle \int_{-1}^{2}\big[3-\ln(y+2)\big],dy$$
$$\displaystyle \int_{-1}^{2}\big[\ln(y+2)-2\big],dy$$
Explanation
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.
Set up the area integral (in $y$) between $x=-y^2+4$ (right) and $x=y-2$ (left) for $0\le y\le 3$.
$\displaystyle \int_{0}^{3}\big[(y-2)-(-y^2+4)\big],dy$
$\displaystyle \int_{0}^{3}\big[(-y^2-4)-(y-2)\big],dy$
$\displaystyle \int_{0}^{3}\big[(-y^2+4)-(y-2)\big],dy$
$\displaystyle \int_{-3}^{0}\big[(-y^2+4)-(y-2)\big],dy$
$\displaystyle \int_{0}^{3}\big[(-y^2+4)-(y+2)\big],dy$
Explanation
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.
Which integral represents the area between $x=4$ (right) and $x=\sin y$ (left) for $0\le y\le \pi$?
$\displaystyle \int_{0}^{\pi}\big[4-\sin y\big]dy$
$\displaystyle \int_{0}^{\pi}\big[4+\sin y\big]dy$
$\displaystyle \int_{0}^{\pi}\big[\sin y-4\big]dy$
$\displaystyle \int_{\pi}^{0}\big[4-\sin y\big]dy$
$\displaystyle \int_{0}^{\pi}\big[4-\cos y\big]dy$
Explanation
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 and the left curve is x = sin y, so the integrand is 4 - sin y. The limits of integration are from y = 0 to y = π, 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.