AP Calculus BC Practice Quiz: Finding the Area Between Curves Expressed as Functions of $y$

Correct Answer: A

To find the area with respect to $y$, we integrate the difference between the right function and the left function. The region is bounded on the right by $x = 4$ and on the left by $x = y^2$. The bounds of integration are along the y-axis. The region starts at $y = 0$. The upper bound is where the curves $x = y^2$ and $x = 4$ intersect, which is $y^2 = 4$, or $y = 2$ in the first quadrant. Therefore, the integral is $\int_{0}^{2} (4 - y^2) dy$.

Correct Answer: B

When calculating the area between curves by integrating with respect to $y$, the definite integral takes the form $\int_{c}^{d} (\text{right curve} - \text{left curve}) dy$. The integrand, $f(y) - g(y)$, represents the horizontal width of a representative rectangle. Thus, $f(y)$ must be the function defining the right boundary (larger x-values) and $g(y)$ must be the function defining the left boundary (smaller x-values).

Correct Answer: A

First, find the points of intersection by setting the functions equal: $y^2 - 2 = y \Rightarrow y^2 - y - 2 = 0 \Rightarrow (y - 2)(y + 1) = 0$. The intersection points occur at $y = -1$ and $y = 2$, which are the limits of integration. To determine the right and left functions, test a point between -1 and 2, such as $y = 0$. For $x = y^2 - 2$, $x = -2$. For $x = y$, $x = 0$. Since $0 > -2$, the line $x = y$ is the right function and the parabola $x = y^2 - 2$ is the left function. The area is given by the integral of (right - left) with respect to $y$, which is $\int_{-1}^{2} (y - (y^2 - 2)) dy$.

Correct Answer: C

The region is being evaluated with respect to $y$. The right boundary of the region is the curve $x = \sqrt{y}$. The left boundary is the y-axis, which has the equation $x = 0$. The region is bounded below by the x-axis ($y=0$) and above by the line $y=4$. Therefore, the limits of integration for $y$ are from 0 to 4. The area is the integral of the (right function - left function) with respect to $y$: $\int_{0}^{4} (\sqrt{y} - 0) dy = \int_{0}^{4} \sqrt{y} \, dy$.

Correct Answer: A

First, find the intersection points: $y + 1 = y^2 - 1 \Rightarrow y^2 - y - 2 = 0 \Rightarrow (y - 2)(y + 1) = 0$. The curves intersect at $y = -1$ and $y = 2$. In the interval $(-1, 2)$, choose a test point like $y=0$. For $x = y+1$, $x=1$. For $x=y^2-1$, $x=-1$. Since $1 > -1$, the line $x=y+1$ is the right function. The area is $\int_{-1}^{2} ((y+1) - (y^2-1)) dy = \int_{-1}^{2} (-y^2 + y + 2) dy$. Evaluating the integral gives $[-\frac{y^3}{3} + \frac{y^2}{2} + 2y]_{-1}^{2} = (-\frac{8}{3} + \frac{4}{2} + 4) - (\frac{1}{3} + \frac{1}{2} - 2) = (-\frac{8}{3} + 6) - (\frac{5}{6} - 2) = \frac{10}{3} - (-\frac{7}{6}) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$.

Correct Answer: B

To find the area of a region by integrating with respect to $y$, we use the definite integral of the difference between the rightmost function and the leftmost function over the interval of y-values that bounds the region. The graph shows that for any $y$ between $c$ and $d$, the x-value on the curve $x=f(y)$ is greater than the x-value on the curve $x=g(y)$. Thus, $f(y)$ is the right boundary and $g(y)$ is the left boundary. The limits of integration are the y-values from $c$ to $d$. Therefore, the correct integral is $\int_{c}^{d} (f(y) - g(y)) dy$.

Correct Answer: C

First, find the intersections: $y^3 = y \Rightarrow y^3 - y = 0 \Rightarrow y(y^2 - 1) = 0$, so $y = -1, 0, 1$. This creates two separate regions. For the interval $[-1, 0]$, test $y = -1/2$: $x = (-1/2)^3 = -1/8$ and $x = -1/2$. Since $-1/8 > -1/2$, $x=y^3$ is the right function. The area for this region is $\int_{-1}^{0} (y^3 - y) dy$. For the interval $[0, 1]$, test $y = 1/2$: $x = (1/2)^3 = 1/8$ and $x = 1/2$. Since $1/2 > 1/8$, $x=y$ is the right function. The area for this region is $\int_{0}^{1} (y - y^3) dy$. The total area is the sum of the areas of the two regions, which must both be positive. Therefore, the total area is $\int_{-1}^{0} (y^3 - y) dy + \int_{0}^{1} (y - y^3) dy$.