Area Between Curves: Functions of y
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AP Calculus AB › Area Between Curves: Functions of y
Find an area setup for region between $x=\sqrt{y+4}$ (right) and $x=\frac{y}{2}$ (left) for $0\le y\le 4$.
$\displaystyle \int_{0}^{4}\left[\frac{y}{2}-\sqrt{y+4}\right]dy$
$\displaystyle \int_{0}^{4}\left[\sqrt{y+4}-\frac{y}{2}\right]dy$
$\displaystyle \int_{4}^{0}\left[\sqrt{y+4}-\frac{y}{2}\right]dy$
$\displaystyle \int_{0}^{4}\left[\sqrt{y+4}-\frac{2}{y}\right]dy$
$\displaystyle \int_{0}^{4}\left[\sqrt{y-4}-\frac{y}{2}\right]dy$
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 + 4) and the left curve is x = y/2, so the integrand is √(y + 4) - y/2. 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.
Find an expression for the area bounded by $x=y^2-2y$ (left) and $x=4-y$ (right) for $-1\le y\le 2$.
$\displaystyle \int_{-1}^{2}\big[(4-y)-(y^2+2y)\big],dy$
$\displaystyle \int_{-1}^{2}\big[(y^2-2y)-(4-y)\big],dy$
$\displaystyle \int_{2}^{-1}\big[(4-y)-(y^2-2y)\big],dy$
$\displaystyle \int_{-1}^{2}\big[(4-y)-(y^2-2y)\big],dy$
$\displaystyle \int_{-1}^{2}\big[(4+y)-(y^2-2y)\big],dy$
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 - 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.
Find the area setup with respect to $y$ between $x=\frac{1}{y+1}$ (right) and $x=y-1$ (left) for $0\le y\le 2$.
$\displaystyle \int_{-2}^{0}\big[\tfrac{1}{y+1}-(y-1)\big],dy$
$\displaystyle \int_{0}^{2}\big[\tfrac{1}{y+1}-(y+1)\big],dy$
$\displaystyle \int_{0}^{2}\big[\tfrac{1}{y-1}-(y-1)\big],dy$
$\displaystyle \int_{0}^{2}\big[(y-1)-\tfrac{1}{y+1}\big],dy$
$\displaystyle \int_{0}^{2}\big[\tfrac{1}{y+1}-(y-1)\big],dy$
Explanation
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.
Which integral represents the area between $x=y^2$ (right) and $x=y^2-4$ (left) for $-2\le y\le 2$?
$\displaystyle \int_{-2}^{2}\big[y^2-(y^2+4)\big]dy$
$\displaystyle \int_{-2}^{2}\big[(y^2-4)-(y^2+4)\big]dy$
$\displaystyle \int_{-2}^{2}\big[(y^2-4)-y^2\big]dy$
$\displaystyle \int_{-2}^{2}\big[y^2-(y^2-4)\big]dy$
$\displaystyle \int_{2}^{-2}\big[y^2-(y^2-4)\big]dy$
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 = $y^2$ 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.
Find an area expression between $x=3-y^2$ (left) and $x=5$ (right) for $-1 \le y \le 1$.
$\displaystyle \int_{-1}^{1}\big[5-(3-y^2)\big]dy$
$\displaystyle \int_{-1}^{1}\big[5-(3-y)\big]dy$
$\displaystyle \int_{1}^{-1}\big[5-(3-y^2)\big]dy$
$\displaystyle \int_{-1}^{1}\big[(3-y^2)-5\big]dy$
$\displaystyle \int_{-1}^{1}\big[5-(3+y^2)\big]dy$
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 $(x_{\text{right}} - x_{\text{left}}) , dy$. 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.
Set up the area between $x=y^3+2$ (right) and $x=y$ (left) on $-1 \le y \le 2$.
$$\displaystyle \int_{-1}^{2}\big[(y^3-2)-y\big]dy$$
$$\displaystyle \int_{2}^{-1}\big[(y^3+2)-y\big]dy$$
$$\displaystyle \int_{-1}^{2}\big[y-(y^3+2)\big]dy$$
$$\displaystyle \int_{-1}^{2}\big[(y^3+2)-y\big]dy$$
$$\displaystyle \int_{-1}^{2}\big[(y^2+2)-y\big]dy$$
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 $x = y^3 + 2$ and the left curve is $x = y$, so the integrand is $(y^3 + 2) - y$. The limits of integration are from $y = -1$ 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.
Set up the area between $x=2\cos y$ (left) and $x=3$ (right) for $-\frac{\pi}{2}\le y\le \frac{\pi}{2}$.
$\displaystyle \int_{\pi/2}^{-\pi/2}\big[3-2\cos y\big]dy$
$\displaystyle \int_{-\pi/2}^{\pi/2}\big[3+2\cos y\big]dy$
$\displaystyle \int_{-\pi/2}^{\pi/2}\big[2\cos y-3\big]dy$
$\displaystyle \int_{-\pi/2}^{\pi/2}\big[3-2\sin y\big]dy$
$\displaystyle \int_{-\pi/2}^{\pi/2}\big[3-2\cos y\big]dy$
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 $x = 3$ 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.
An integral for the area between $x=\frac{1}{y+1}$ (right) and $x=0$ (left) on $1\le y\le 3$ is
$\displaystyle \int_{3}^{1}\left[\frac{1}{y+1}-0\right]dy$
$\displaystyle \int_{1}^{3}\left[\frac{1}{1-y}-0\right]dy$
$\displaystyle \int_{1}^{3}\left[\frac{1}{y-1}-0\right]dy$
$\displaystyle \int_{1}^{3}\left[0-\frac{1}{y+1}\right]dy$
$\displaystyle \int_{1}^{3}\left[\frac{1}{y+1}-0\right]dy$
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 = 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.
Which integral gives area between $x=2y^2$ (left) and $x=8$ (right) for $-2\le y\le 2$?
$\displaystyle \int_{-2}^{2}\big[8-(2y^2)\big]dy$
$\displaystyle \int_{-2}^{2}\big[(2y^2)-8\big]dy$
$\displaystyle \int_{2}^{-2}\big[8-(2y^2)\big]dy$
$\displaystyle \int_{-2}^{2}\big[8-(2y)\big]dy$
$\displaystyle \int_{-2}^{2}\big[8-(y^2)\big]dy$
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.
Which integral represents the area between $x=\sqrt{y}$ (left) and $x=\frac{y}{2}+1$ (right) for $0\le y\le 4$?
$\displaystyle \int_{0}^{4}\left[\left(\frac{y}{2}-1\right)-\sqrt{y}\right]dy$
$\displaystyle \int_{0}^{4}\left[\left(\frac{y}{2}+1\right)-\sqrt{y+1}\right]dy$
$\displaystyle \int_{0}^{4}\left[\sqrt{y}-\left(\frac{y}{2}+1\right)\right]dy$
$\displaystyle \int_{4}^{0}\left[\left(\frac{y}{2}+1\right)-\sqrt{y}\right]dy$
$\displaystyle \int_{0}^{4}\left[\left(\frac{y}{2}+1\right)-\sqrt{y}\right]dy$
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.