Area Between Curves (y-functions) - Calculus 1

1. What is the total area of the region bounded by $x=y\sqrt{1-y^2}$ and the y-axis?
   Answer: $\frac{2}{3}$ The domain of the function $x=y\sqrt{1-y^2}$ is where $1-y^2 \ge 0$, which is $-1 \le y \le 1$. The curve intersects the y-axis ($x=0$) at $y=0, y=1, y=-1$. We need to find the total area, so we must consider the sign of $x$. For $y \in (0, 1)$, $x > 0$, so $x=y\sqrt{1-y^2}$ is the right curve. For $y \in (-1, 0)$, $x < 0$, so $x=0$ is the right curve. The total area is $A = \int_{-1}^{0} (0 - y\sqrt{1-y^2}) \,dy + \int_{0}^{1} (y\sqrt{1-y^2} - 0) \,dy$. Notice the integrand is an odd function of $y$, so the regions have equal area. We can compute one integral and double it: $A = 2 \int_{0}^{1} y\sqrt{1-y^2} \,dy$. Use u-substitution with $u = 1-y^2$, so $du = -2y \,dy$, or $y\,dy = -du/2$. The limits change: when $y=0, u=1$; when $y=1, u=0$. The integral becomes $A = 2 \int_{1}^{0} \sqrt{u} (-\frac{du}{2}) = -\int_{1}^{0} u^{1/2} \,du = \int_{0}^{1} u^{1/2} \,du = [\frac{2}{3}u^{3/2}]_0^1 = \frac{2}{3}$.
2. Consider the region in the first quadrant enclosed by the y-axis and the curve $x = 9 - y^2$. What is the area of the portion of this region for which $x \ge 5$?
   Answer: $\frac{16}{3}$ We are looking for the area in the first quadrant, so $y \ge 0$. The region is bounded by $x=9-y^2$ and is restricted to where $x \ge 5$. This means the left boundary is the line $x=5$. To find the y-limits, we find the intersection of $x=9-y^2$ and $x=5$, which is $5 = 9-y^2 \implies y^2=4 \implies y=\pm 2$. Since we are in the first quadrant, the y-limits are from 0 to 2. The area is $A = \int_{0}^{2} ((9 - y^2) - 5) \,dy = \int_{0}^{2} (4 - y^2) \,dy$. Evaluating the integral: $A = [4y - \frac{y^3}{3}]_0^2 = (4(2) - \frac{2^3}{3}) - 0 = 8 - \frac{8}{3} = \frac{16}{3}$.
3. The area between the curves $x=f(y)$ and $x=g(y)$ from $y=a$ to $y=b$ is correctly given by the integral $\int_a^b |f(y)-g(y)| \,dy$. Which of the following conditions is NOT strictly necessary for this formula to be valid?
   Answer: The curves $x=f(y)$ and $x=g(y)$ must intersect at $y=a$ and $y=b$. The formula for the area between two curves $x=f(y)$ and $x=g(y)$ on an interval $[a, b]$ requires the functions to be continuous so the integral is well-defined. The use of absolute value ensures the integrand $|f(y)-g(y)|$ is non-negative, correctly representing the distance between the curves, and thus the integral gives a non-negative total area. The limits of integration $a$ and $b$ define the interval, and it is standard convention that $a \le b$. However, it is not necessary for the curves to intersect at the endpoints of the interval. The region can be bounded by the lines $y=a$ and $y=b$ without the curves intersecting at these y-values.
4. A region is bounded by $x = \frac{1}{y}$, $x=0$, $y=1$, and $y=b$ where $b>1$. If the area of this region is 2, what is the value of $b$?
   Answer: $e^2$ When you encounter a region bounded by curves and need to find an unknown boundary given the area, you're dealing with definite integration where the limits or integrand contain a parameter. First, let's visualize this region. You have $x = \frac{1}{y}$ (a hyperbola), the y-axis ($x = 0$), and horizontal lines at $y = 1$ and $y = b$. Since we're integrating with respect to $y$, the area is: $\text{Area} = \int_1^b \left(\frac{1}{y} - 0\right) dy = \int_1^b \frac{1}{y} dy$ The antiderivative of $\frac{1}{y}$ is $\ln|y|$, so: $\text{Area} = [\ln|y|]_1^b = \ln(b) - \ln(1) = \ln(b)$ Since the area equals 2: $\ln(b) = 2$, which means $b = e^2$. Choice A ($e$) would give an area of $\ln(e) = 1$, which is too small. Choice B (3) would give an area of $\ln(3) \approx 1.1$, still too small. Choice C ($2e$) would give an area of $\ln(2e) = \ln(2) + \ln(e) = \ln(2) + 1 \approx 1.69$, which is close but not exactly 2. Only choice D ($e^2$) satisfies our equation since $\ln(e^2) = 2\ln(e) = 2(1) = 2$. Study tip: When solving area problems with unknown boundaries, set up your integral first, then use the given area as an equation to solve for the unknown parameter. Remember that $\ln(e^n) = n$ is a crucial relationship for exponential-logarithmic problems.
5. What is the area of the region bounded by $x = |y-1|$ and $x = (y-1)^2$?
   Answer: $\frac{1}{3}$ When finding the area between two curves, you need to determine which curve is the outer boundary and set up your integral carefully. This problem involves absolute value, so start by analyzing the behavior of both functions. The curve $x = |y-1|$ creates a V-shape opening to the right with vertex at $(0,1)$. For $y \geq 1$, this becomes $x = y-1$, and for $y \leq 1$, it becomes $x = -(y-1) = 1-y$. The parabola $x = (y-1)^2$ also has vertex at $(0,1)$ but opens to the right. To find intersection points, solve $|y-1| = (y-1)^2$. Let $u = y-1$, so $|u| = u^2$. This gives us $u = 0$ (where $y = 1$) and $u = 1$ (where $y = 2$), and $u = -1$ (where $y = 0$). The intersection points are $(0,0)$, $(0,1)$, and $(1,2)$. Between $y = 0$ and $y = 2$, the absolute value function lies to the right of the parabola. The area is: $\int_0^2 [|y-1| - (y-1)^2] dy = \int_0^1 [(1-y) - (y-1)^2] dy + \int_1^2 [(y-1) - (y-1)^2] dy$ Evaluating these integrals gives $\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$. Answer choice A ($\frac{1}{6}$) represents only half the region. Answer B ($1$) likely comes from incorrect setup or integration bounds. Answer C ($\frac{2}{3}$) suggests doubling the correct answer, possibly from confusion about symmetry. Remember: with absolute value problems, always split the integral at the point where the expression inside changes sign, and carefully track which function is the outer boundary in each interval.
6. A region is bounded by the curve $x=4-y^2$ and the line $x=-5$. Which of the following integrals represents the area of this region?
   Answer: $\int_{-3}^{3} (9-y^2) dy$ When finding the area between curves, you need to carefully identify the region's boundaries and set up your integral with the correct limits and integrand. First, let's visualize this region. The curve $x = 4 - y^2$ is a parabola opening leftward with vertex at (4, 0). The line $x = -5$ is vertical. To find where they could intersect, set $4 - y^2 = -5$, giving $y^2 = 9$, so $y = \pm 3$. However, since the parabola has its rightmost point at $x = 4$ and the line is at $x = -5$, they don't actually intersect. The region is bounded by the parabola on the right and extends leftward to the line $x = -5$. For the area calculation using horizontal strips (integrating with respect to $y$), the width of each strip runs from $x = -5$ to $x = 4 - y^2$. This gives a width of $(4 - y^2) - (-5) = 9 - y^2$. The parabola extends from $y = -3$ to $y = 3$, so the area is $\int_{-3}^{3} (9 - y^2) dy$. This matches choice A. Choice B uses the wrong limits ($-2$ to 2 instead of $-3$ to 3). Choice C attempts integration with respect to $x$, but the setup is incorrect for this region. Choice D has the wrong integrand ($-1 - y^2$ instead of $9 - y^2$), which would give a negative area. Strategy tip: Always sketch the region first, identify intersection points to determine limits, and ensure your integrand represents the actual distance between boundaries.
7. What is the area of the region bounded by the curves $x = e^y$, $x = 1/e$, and the lines $y = -1$ and $y = 1$?
   Answer: $e - \frac{3}{e}$ The region is defined by the vertical boundaries $y=-1$ and $y=1$. We need to determine the right and left curves, $x_{right}$ and $x_{left}$. The curves are $x = e^y$ and $x=1/e$. Note that at $y=-1$, $e^y = e^{-1} = 1/e$. For $y > -1$, $e^y > 1/e$. Therefore, $x=e^y$ is the right curve throughout the interval $[-1, 1]$. The area is given by $A = \int_{-1}^{1} (e^y - \frac{1}{e}) \,dy$. Evaluating the integral: $A = [e^y - \frac{y}{e}]_{-1}^{1} = (e^1 - \frac{1}{e}) - (e^{-1} - \frac{-1}{e}) = (e - \frac{1}{e}) - (\frac{1}{e} + \frac{1}{e}) = e - \frac{1}{e} - \frac{2}{e} = e - \frac{3}{e}$.
8. The area of the region bounded by the parabola $x = ay^2$ (for $a>0$) and the line $x=4$ is $64/3$. What is the value of the constant $a$?
   Answer: $\frac{1}{4}$ The region is bounded by $x=4$ (right curve) and $x=ay^2$ (left curve). Find the intersection points by setting $ay^2 = 4$, which gives $y^2 = 4/a$, so $y = \pm 2/\sqrt{a}$. These are the limits of integration. The area is $A = \int_{-2/\sqrt{a}}^{2/\sqrt{a}} (4 - ay^2) \,dy$. Using symmetry, $A = 2 \int_{0}^{2/\sqrt{a}} (4 - ay^2) \,dy$. Evaluating the integral: $A = 2[4y - \frac{ay^3}{3}]_0^{2/\sqrt{a}} = 2(4(\frac{2}{\sqrt{a}}) - \frac{a}{3}(\frac{8}{a\sqrt{a}})) = 2(\frac{8}{\sqrt{a}} - \frac{8}{3\sqrt{a}}) = 2(\frac{16}{3\sqrt{a}}) = \frac{32}{3\sqrt{a}}$. We are given that the area is $64/3$. So, $\frac{32}{3\sqrt{a}} = \frac{64}{3} \implies 32 = 64\sqrt{a} \implies \sqrt{a} = \frac{32}{64} = \frac{1}{2}$. Squaring both sides gives $a = \frac{1}{4}$.
9. A region's area is given by the integral $\int_{-1}^{3} (2y+3 - y^2) \,dy$. Which pair of curves bounds this region?
   Answer: $x=y^2$ and $x=2y+3$ The integral is of the form $\int_c^d (x_{right} - x_{left}) \,dy$. The integrand is $(2y+3) - y^2$. This suggests that the right curve is $x_{right} = 2y+3$ and the left curve is $x_{left} = y^2$. To verify this, we check if the limits of integration, $y=-1$ and $y=3$, are the intersection points of these two curves. Set $y^2 = 2y+3$, which gives $y^2 - 2y - 3 = 0$. This factors as $(y-3)(y+1) = 0$, yielding $y=3$ and $y=-1$. These match the limits of integration. Thus, the curves are $x=y^2$ and $x=2y+3$.
10. A region in the xy-plane is bounded by $x = y^2 + 1$, $x = 5$, $y = -1$, and $y = 2$. What is the area of this region?
    Answer: $9$ The integration is with respect to $y$ from $y=-1$ to $y=2$. We must determine the right and left boundary curves. The boundaries are $x = y^2 + 1$ and $x=5$. The curves intersect when $y^2+1=5$, which is at $y=\pm 2$. Since the interval of integration is $[-1, 2]$, and for any $y$ in this interval $y^2+1 \le 5$, the line $x=5$ is the right boundary and the parabola $x=y^2+1$ is the left boundary. The area is $A = \int_{-1}^{2} (5 - (y^2 + 1)) \,dy = \int_{-1}^{2} (4 - y^2) \,dy$. Evaluating the integral: $A = [4y - \frac{y^3}{3}]_{-1}^{2} = (4(2) - \frac{2^3}{3}) - (4(-1) - \frac{(-1)^3}{3}) = (8 - \frac{8}{3}) - (-4 + \frac{1}{3}) = \frac{16}{3} - (-\frac{11}{3}) = \frac{16+11}{3} = \frac{27}{3} = 9$.
11. A region is bounded by the curves $x = y^2 - 2$ and $x = |y|$. What is the area of this region?
    Answer: $\frac{20}{3}$ To find the points of intersection, we solve $y^2 - 2 = |y|$. For $y \ge 0$, $y^2 - 2 = y \implies y^2 - y - 2 = 0 \implies (y-2)(y+1)=0$, so $y=2$. For $y < 0$, $y^2 - 2 = -y \implies y^2 + y - 2 = 0 \implies (y+2)(y-1)=0$, so $y=-2$. The limits of integration are from $y=-2$ to $y=2$. In the interval $(-2, 2)$, test $y=0$: $x = -2$ and $x = 0$. Thus, $x=|y|$ is the right curve. The area is $A = \int_{-2}^{2} (|y| - (y^2 - 2)) \,dy$. Due to symmetry of the integrand, we can write $A = 2 \int_{0}^{2} (y - (y^2 - 2)) \,dy = 2 \int_{0}^{2} (y - y^2 + 2) \,dy$. Evaluating the integral: $2[\frac{y^2}{2} - \frac{y^3}{3} + 2y]_0^2 = 2[(\frac{4}{2} - \frac{8}{3} + 4) - 0] = 2(6 - \frac{8}{3}) = 2(\frac{10}{3}) = \frac{20}{3}$.
12. What is the area of the region bounded by $x = 2\cos(y)$, $x = \cos(y)$, $y=0$, and $y=\pi/2$?
    Answer: $1$ The region is bounded by the y-values 0 and $\pi/2$. In the interval $y \in [0, \pi/2]$, $\cos(y) \ge 0$. Therefore, $2\cos(y) \ge \cos(y)$, so $x = 2\cos(y)$ is the right curve. The area is given by the integral $A = \int_{0}^{\pi/2} (2\cos(y) - \cos(y)) \,dy = \int_{0}^{\pi/2} \cos(y) \,dy$. The antiderivative of $\cos(y)$ is $\sin(y)$. Evaluating the definite integral: $A = [\sin(y)]_0^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1$.
13. Which of the following integrals represents the area of the region enclosed by the curves defined by the equations $x + y^2 = 3$ and $x + y = 1$?
    Answer: $\int_{-1}^{2} (2 + y - y^2) \,dy$ First, express $x$ in terms of $y$ for both equations: $x = 3 - y^2$ and $x = 1 - y$. To find the limits of integration, set the expressions for $x$ equal to each other: $3 - y^2 = 1 - y$, which simplifies to $y^2 - y - 2 = 0$, or $(y-2)(y+1) = 0$. The points of intersection occur at $y = -1$ and $y = 2$. These are the limits of integration. To determine which function is on the right, test a value between -1 and 2, such as $y=0$. For $y=0$, $x = 3 - 0^2 = 3$ and $x = 1 - 0 = 1$. Since $3 > 1$, the curve $x = 3 - y^2$ is the right boundary ($x_{right}$) and $x = 1 - y$ is the left boundary ($x_{left}$). The area is given by the integral $A = \int_{c}^{d} (x_{right} - x_{left}) \,dy$. Therefore, the area is $\int_{-1}^{2} ((3 - y^2) - (1 - y)) \,dy = \int_{-1}^{2} (2 + y - y^2) \,dy$.
14. Find the area of the region enclosed by the graphs of the equations $y = x - 1$ and $y^2 = 2x + 6$.
    Answer: $18$ To find the area by integrating with respect to $y$, we first express $x$ as a function of $y$ for both equations. From $y = x - 1$, we get $x = y + 1$. From $y^2 = 2x + 6$, we get $2x = y^2 - 6$, so $x = \frac{1}{2}y^2 - 3$. Next, we find the points of intersection by setting the expressions for $x$ equal: $y + 1 = \frac{1}{2}y^2 - 3$. Multiplying by 2 gives $2y + 2 = y^2 - 6$, which simplifies to $y^2 - 2y - 8 = 0$. Factoring gives $(y - 4)(y + 2) = 0$, so the curves intersect at $y = -2$ and $y = 4$. These are our limits of integration. To determine which function is on the right, we can test a value of $y$ between $-2$ and $4$, such as $y=0$. For $x = y + 1$, we get $x=1$. For $x = \frac{1}{2}y^2 - 3$, we get $x=-3$. Thus, $x = y + 1$ is the right curve. The area $A$ is given by the integral of the right function minus the left function: $A = \int_{-2}^{4} \left( (y+1) - (\frac{1}{2}y^2 - 3) \right) dy = \int_{-2}^{4} \left( -\frac{1}{2}y^2 + y + 4 \right) dy$ $A = \left[ -\frac{1}{6}y^3 + \frac{1}{2}y^2 + 4y \right]_{-2}^{4}$ $A = \left( -\frac{64}{6} + \frac{16}{2} + 16 \right) - \left( -\frac{-8}{6} + \frac{4}{2} - 8 \right) = \left( -\frac{32}{3} + 8 + 16 \right) - \left( \frac{4}{3} + 2 - 8 \right)$ $A = \left( \frac{40}{3} \right) - \left( -\frac{14}{3} \right) = \frac{54}{3} = 18$ The distractor $36$ comes from an algebraic error when solving for $x$ from $y^2=2x+6$, specifically forgetting to divide the constant term by 2, leading to an incorrect integrand of $-\frac{1}{2}y^2+y+7$. The distractor $27$ can arise from a sign error when evaluating the definite integral. The distractor $12$ may result from incorrect bounds of integration.
15. Let R be the region bounded by $x = \sin(\pi y)$ and $x = 2y$ for $0 \le y \le 1/2$. Which integral represents the area of R?
    Answer: $\int_{0}^{1/2} (\sin(\pi y) - 2y) \,dy$ The problem specifies the interval of integration for $y$ is $[0, 1/2]$. We need to determine which function is greater on this interval. The curves intersect at $y=0$ since $\sin(0)=0$ and $2(0)=0$. They also intersect at $y=1/2$ since $\sin(\pi/2)=1$ and $2(1/2)=1$. To find which function is greater between these points, we can check the concavity of $f(y) = \sin(\pi y) - 2y$. $f''(y) = -\pi^2 \sin(\pi y)$, which is negative for $y \in (0, 1/2)$. A function that is zero at the endpoints and concave down in between must be positive. Therefore, $\sin(\pi y) - 2y > 0$, which means $\sin(\pi y) > 2y$ on $(0, 1/2)$. So $x=\sin(\pi y)$ is the right curve. The area is $A = \int_{0}^{1/2} (\sin(\pi y) - 2y) \,dy$.
16. The total area of the region enclosed by the curve $x = y^3 - 4y$ and the y-axis ($x=0$) is to be calculated. Which statement correctly describes the setup?
    Answer: The area is given by $\int_{-2}^{0} (y^3 - 4y) \,dy + \int_{0}^{2} (0 - (y^3 - 4y)) \,dy$. First, find the intersections with the y-axis by setting $x=0$: $y^3 - 4y = 0 \implies y(y^2-4)=0$, which gives $y=0, y=2, y=-2$. The function $x=f(y)=y^3-4y$ changes sign in the interval $[-2,2]$. For $y \in (-2, 0)$, let $y=-1$, then $x = (-1)^3 - 4(-1) = 3 > 0$, so $x=y^3-4y$ is the right curve. For $y \in (0, 2)$, let $y=1$, then $x = 1^3 - 4(1) = -3 < 0$, so $x=0$ is the right curve. To find the total area, we must split the integral where the right and left curves switch. This leads to two integrals: $A = \int_{-2}^{0} (y^3 - 4y - 0) \,dy + \int_{0}^{2} (0 - (y^3 - 4y)) \,dy$. Option A would yield 0, which is incorrect for area. Option C incorrectly assumes the integrand is positive on $[0,2]$ and misapplies symmetry. Option D is false.
17. Find the area of the region bounded by the curve $x = \tan(y)$, the y-axis, and the line $y = \pi/3$.
    Answer: $\ln(2)$ The region is bounded by $x=\tan(y)$ (right curve, since $\tan(y) \ge 0$ for $y \in [0, \pi/3]$), $x=0$ (the y-axis, left curve), and the line $y=\pi/3$. The lower bound is implicitly $y=0$ since that's where $x=\tan(y)$ intersects the y-axis. The area is $A = \int_{0}^{\pi/3} (\tan(y) - 0) \,dy$. The antiderivative of $\tan(y)$ is $-\ln|\cos(y)|$. Evaluating the definite integral: $A = [-\ln|\cos(y)|]_0^{\pi/3} = -\ln|\cos(\pi/3)| - (-\ln|\cos(0)|) = -\ln(1/2) + \ln(1)$. Since $\ln(1)=0$ and $-\ln(1/2) = -(\ln(1) - \ln(2)) = \ln(2)$, the area is $\ln(2)$.
18. Find the area of the region bounded by the curve $x=y^2-2y$ and the $y$-axis.
    Answer: $\frac{4}{3}$ When finding areas bounded by curves, you need to set up the correct integral with proper limits and integrand. Since the curve is given as $x = y^2 - 2y$ (a function of $y$), you should integrate with respect to $y$. First, find where the curve intersects the $y$-axis by setting $x = 0$: $0 = y^2 - 2y = y(y-2)$. This gives $y = 0$ and $y = 2$ as your integration limits. To visualize this region, complete the square: $x = y^2 - 2y = (y-1)^2 - 1$. This parabola opens rightward with vertex at $(-1, 1)$. Between $y = 0$ and $y = 2$, the curve lies to the left of the $y$-axis (since $x < 0$ in this interval), so the area is: $\text{Area} = \int_0^2 |y^2 - 2y| \, dy = \int_0^2 -(y^2 - 2y) \, dy = \int_0^2 (2y - y^2) \, dy$ Evaluating: $\int_0^2 (2y - y^2) \, dy = \left[y^2 - \frac{y^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$ So A is correct. Option B ($\frac{2}{3}$) likely comes from forgetting the absolute value or making an arithmetic error. Option C ($2$) might result from incorrectly using $\int_0^2 2y \, dy$ without the $-y^2$ term. Option D ($\frac{8}{3}$) could come from using $\int_0^2 (y^2 - 2y) \, dy$ without taking the absolute value. Remember: when integrating to find area, always use absolute value to ensure positive results, and choose your variable of integration based on how the curve is parameterized.
19. What is the area of the finite region bounded by the curves $x = 2y^2$ and $x = 3y + 2$?
    Answer: $\frac{125}{24}$ First, find the intersection points by setting the expressions for $x$ equal: $2y^2 = 3y + 2$. This gives the quadratic equation $2y^2 - 3y - 2 = 0$. Factoring yields $(2y + 1)(y - 2) = 0$, so the intersection points occur at $y = -1/2$ and $y = 2$. These are the limits of integration. To determine the right and left functions, test a value between the limits, like $y=0$. For the parabola $x = 2y^2$, $x=0$. For the line $x = 3y + 2$, $x=2$. The line is the right-hand boundary. The area is given by the integral: $A = \int_{-1/2}^{2} ((3y + 2) - 2y^2) dy$ $A = \left[ \frac{3}{2}y^2 + 2y - \frac{2}{3}y^3 \right]_{-1/2}^{2}$ Evaluate at the upper limit: $\frac{3}{2}(2)^2 + 2(2) - \frac{2}{3}(2)^3 = 6 + 4 - \frac{16}{3} = 10 - \frac{16}{3} = \frac{14}{3}$. Evaluate at the lower limit: $\frac{3}{2}(-\frac{1}{2})^2 + 2(-\frac{1}{2}) - \frac{2}{3}(-\frac{1}{2})^3 = \frac{3}{8} - 1 - \frac{2}{3}(-\frac{1}{8}) = \frac{3}{8} - 1 + \frac{1}{12} = \frac{9-24+2}{24} = -\frac{13}{24}$. The area is the difference: $A = \frac{14}{3} - (-\frac{13}{24}) = \frac{112}{24} + \frac{13}{24} = \frac{125}{24}$. Distractor D arises from incorrectly factoring the quadratic as $(2y-1)(y+2)=0$, leading to wrong integration limits. Distractor B arises from an error in the antiderivative (e.g., forgetting the $1/2$ in $\frac{3}{2}y^2$). Distractor A results from an arithmetic error in combining the evaluated limits, such as $\frac{112}{24} - \frac{13}{24}$ instead of adding.
20. Find the area of the region enclosed by the curves $x = y^4$ and $x = 8y$.
    Answer: $\frac{48}{5}$ Find the intersection points by setting $y^4 = 8y$. This gives $y^4 - 8y = 0$, or $y(y^3 - 8) = 0$. The solutions are $y=0$ and $y=2$. These are the limits of integration. For $y$ in the interval $(0, 2)$, choose a test point like $y=1$. We have $x = 1^4 = 1$ and $x = 8(1) = 8$. Since $8 > 1$, the line $x=8y$ is the right curve. The area is $A = \int_{0}^{2} (8y - y^4) \,dy$. Evaluate the integral: $A = [4y^2 - \frac{y^5}{5}]_0^2 = (4(2^2) - \frac{2^5}{5}) - 0 = (16 - \frac{32}{5}) = \frac{80 - 32}{5} = \frac{48}{5}$.

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