📝 Area between curves integrating with respect to y (35 MCQs)

📖 Explanation: To integrate with respect to $y$, we rewrite both equations as functions of $y$: $x = y^2$ and $x = y+2$. Setting them equal: $y^2 = y+2$ gives $y^2 - y - 2 = 0$, so $(y-2)(y+1)=0$, yielding $y = -1$ and $y = 2$. Over this interval, $y+2 \geq y^2$, so $x = y+2$ is the right boundary. Option A correctly identifies both the intersection points and the right curve.

📖 Explanation: The region is symmetric about the x-axis. The curve $x = 4 - y^2$ intersects the y-axis where $x=0$, so $4 - y^2 = 0$ gives $y = \pm 2$. To find the total area, we can integrate from $y=0$ to $y=2$ and double the result. The horizontal cross-section from the y-axis to the curve has length $4 - y^2$. Thus the area is $2\int_{0}^{2} (4 - y^2) \, dy$. Option D correctly accounts for symmetry, while option A integrates over the full interval but misses the factor of 2 needed for the symmetric extension.

📖 Explanation: Both curves are already expressed as functions of $y$: $x = y^2 - 4$ and $x = 4 - y^2$. Finding their intersections: $y^2 - 4 = 4 - y^2$ gives $2y^2 = 8$, so $y = \pm 2$. The right curve is $x = 4 - y^2$ and the left is $x = y^2 - 4$. Integrating with respect to $y$ over $[-2,2]$ gives the area directly without splitting the region. Option B is the most efficient method. Option A would require solving for y and possibly splitting the region, making it more complex.

📖 Explanation: The student's integral $\int_{-2}^{2} (4 - y^2) \, dy$ actually gives the area between the curve and the y-axis from $y=-2$ to $y=2$. This is correct as written because the curve $x = y^2$ intersects $x=4$ at $y = \pm 2$, and the region between them has horizontal cross-sections of length $4 - y^2$. However, the student may have overlooked that the region is symmetric and could be computed as $2\int_{0}^{2} (4 - y^2) \, dy$. The integral as written is correct, so the real error would be in the limits if they mistakenly used $0$ to $4$ or the wrong integrand. Option A suggests a common misconception about symmetry but the original integral is actually correct for the full region.

📖 Explanation: The curves $x = \sqrt{y}$ and $x = y^2$ intersect when $\sqrt{y} = y^2$. Squaring both sides: $y = y^4$, so $y(1 - y^3) = 0$, giving $y = 0$ and $y = 1$. For $0 < y < 1$, $\sqrt{y} > y^2$, so the right curve is $x = \sqrt{y}$ and the left curve is $x = y^2$. Thus the area is $\int_{0}^{1} (\sqrt{y} - y^2) \, dy$. Option A is correct. Option B reverses the order, option C uses incorrect limits, and option D uses the wrong second function.

📖 Explanation: The curve $x = y^2$ is a parabola opening to the right. The line $y = 2$ is horizontal. To find the area between the curve and the y-axis from $y = -2$ to $y = 2$, we use horizontal cross-sections. The right boundary is $x = y^2$ and the left boundary is the y-axis ($x = 0$). However, if the region is bounded by $y = 2$ and the y-axis, and the curve $x = y^2$, the limits of integration should be from $-2$ to $2$ because the curve intersects $y=2$ at $x=4$ and the y-axis at $y=0$. The integrand is $y^2 - 0 = y^2$. But if we want the full region, we integrate from $y=-2$ to $y=2$ with integrand $4 - y^2$ only if the right boundary is $x=4$. The correct setup for the region enclosed by $x = y^2$, $y=2$, and the y-axis is $\int_{0}^{2} y^2 \, dy$ if we consider only the first quadrant. However, the question likely refers to the symmetric region, so $2\int_{0}^{2} y^2 \, dy$. Option D is incorrect as it introduces $4 - y^2$, which would be for a different region.

📖 Explanation: The integral $\int_{-2}^{2} (4 - y^2) \, dy$ represents the area between the parabola $x = y^2$ and the vertical line $x = 4$. Because the parabola is symmetric about the x-axis, the region from $y = -2$ to $y = 2$ includes both the upper and lower halves. The integral of $(4 - y^2)$ from $-2$ to $2$ gives the total area, which is exactly twice the area in the first quadrant (from $y=0$ to $y=2$). Thus, the student's setup is correct and represents the full region. Option B correctly identifies that this integral is twice the first-quadrant area. The other options misinterpret the region represented by the integral.

📖 Explanation: The curves $x = y^3$ and $x = y$ intersect when $y^3 = y$, so $y(y^2 - 1) = 0$, giving $y = -1, 0, 1$. For $0 \le y \le 1$, we have $y \ge y^3$, so the right curve is $x = y$ and the left curve is $x = y^3$. The region is symmetric about the origin, so the total area is $2\int_{0}^{1} (y - y^3) \, dy$. However, if the question asks for the area between $y = -1$ and $y = 1$, the integral $\int_{-1}^{1} (y - y^3) \, dy$ would be zero because the function is odd. The correct area is $2\int_{0}^{1} (y - y^3) \, dy = \int_{0}^{1} (y - y^3) \, dy$ for the first quadrant. Option A is the correct setup for the area in the first quadrant. Option D would give zero due to symmetry.

📖 Explanation: The curves $x = y^2$ and $x = 2y$ intersect when $y^2 = 2y$, so $y(y - 2) = 0$, giving $y = 0$ and $y = 2$. For $0 < y < 2$, $2y > y^2$, so the right curve is $x = 2y$ and the left curve is $x = y^2$. The area is $\int_{0}^{2} (2y - y^2) \, dy$. Evaluating: $\left[y^2 - \frac{y^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$. Option C is correct. Option B reverses the integrand, which would give a negative area. Options A and D are incorrectly formatted.

📖 Explanation: The curves $x = \sin y$ and $x = \cos y$ intersect when $\sin y = \cos y$, which occurs at $y = \pi/4$. For $0 \le y \le \pi/4$, $\cos y \ge \sin y$, so the right curve is $x = \cos y$ and the left is $x = \sin y$. For $\pi/4 \le y \le \pi/2$, $\sin y \ge \cos y$, so the right curve is $x = \sin y$ and the left is $x = \cos y$. Thus the area must be split into two integrals: $\int_{0}^{\pi/4} (\cos y - \sin y) \, dy + \int_{\pi/4}^{\pi/2} (\sin y - \cos y) \, dy$. Option B correctly splits the integral at the intersection point. Option D uses absolute value but would require the same splitting to evaluate.

📖 Explanation: The curves $x = y^2$ and $x = 2 - y^2$ intersect when $y^2 = 2 - y^2$, so $2y^2 = 2$, giving $y = \pm 1$. The student's integral uses limits $-\sqrt{2}$ to $\sqrt{2}$, which are the x-intercepts of the second curve, not the intersection points. The correct limits should be $-1$ to $1$. The integrand $2 - 2y^2$ is correct for the difference between the curves. However, the limits are wrong because the region enclosed by the two curves only exists between $y = -1$ and $y = 1$. The student's integral would include area where the curves do not enclose a region. Option D incorrectly suggests the integral represents the enclosed area, but it actually represents the area under the difference function over a larger interval that includes points outside the enclosed region.

📖 Explanation: The curve $x = y^2 + 1$ is a parabola opening to the right. The line $x = 5$ is vertical. They intersect when $y^2 + 1 = 5$, so $y^2 = 4$, giving $y = \pm 2$. The region between the parabola and the line extends from $y = -2$ to $y = 2$. The horizontal cross-sections have length $5 - (y^2 + 1) = 4 - y^2$. Thus the limits of integration are $-2$ to $2$. Option B is correct. Option A only considers the upper half, option C confuses x-values with y-limits, and option D extends beyond the intersection points.

📖 Explanation: The curve $x = 4y - y^2$ is a parabola opening to the left. It intersects the y-axis when $x = 0$, so $4y - y^2 = 0$, giving $y(4 - y) = 0$, so $y = 0$ and $y = 4$. The region between the curve and the y-axis extends from $y = 0$ to $y = 4$. The horizontal cross-sections have length $4y - y^2$. The area is $\int_{0}^{4} (4y - y^2) \, dy$. Option A is correct. Options B, C, and D are essentially the same but phrased differently; the key is that the integrand is positive on $(0,4)$, so it correctly represents the area.

📖 Explanation: The curve $x = y^3 - 4y = y(y^2 - 4) = y(y-2)(y+2)$. It intersects the y-axis when $x=0$, so $y = -2, 0, 2$. The function is positive on $(-2,0)$ and negative on $(0,2)$, so the absolute value is needed to get the total area. The integral $\int_{-2}^{2} |y^3 - 4y| \, dy$ gives the total area. Option A would give zero because the function is odd. Option B and C incorrectly split the integral by taking the positive parts but may miss the correct sign. Option D is the correct approach using absolute value, though it's more Hard because it requires splitting at the zeros: $\int_{-2}^{0} (y^3 - 4y) \, dy + \int_{0}^{2} (4y - y^3) \, dy$.

📖 Explanation: The integral $\int_{0}^{2} (y^2 - 2y) \, dy$ has integrand $y^2 - 2y$, which is negative for $0 < y < 2$. This suggests that the right curve is $x = y^2$ and the left curve is $x = 2y$, but since $2y > y^2$ on $(0,2)$, the correct integrand should be $2y - y^2$. Thus the integral as written would give a negative area, indicating a setup error. The curves do intersect at $y=0$ and $y=2$, which is true regardless of the sign. Option C is the only statement that is definitively correct without error. Options A and B incorrectly identify the right curve, and option D is incorrect because the integral can be correct if the limits or integrand are adjusted.

📖 Explanation: The curves intersect when $3y - y^2 = 2y - 2$, so $-y^2 + y + 2 = 0$, or $y^2 - y - 2 = 0$, giving $y = -1$ and $y = 2$. For $-1 < y < 2$, $3y - y^2 > 2y - 2$ because $y - y^2 + 2 > 0$ (verify at $y=0$: $2 > -2$). So the right curve is $x = 3y - y^2$ and the left is $x = 2y - 2$. The integrand is $3y - y^2 - (2y - 2) = 3y - y^2 - 2y + 2 = y - y^2 + 2$. Thus the area is $\int_{-1}^{2} (y - y^2 + 2) \, dy$. Option D is the simplified correct integrand. Option A is the unsimplified correct form. Options B and C have the wrong sign or order.

📖 Explanation: When both curves are naturally expressed as functions of $y$ (i.e., $x = f(y)$), the most straightforward method is to integrate with respect to $y$. The formula $\int_{c}^{d} (w(y) - v(y)) \, dy$ applies where $w(y)$ is the right boundary and $v(y)$ is the left boundary. Option B correctly identifies this method. Option A would require solving for y in terms of x, which may be more difficult or introduce multiple branches. Options C and D are methods for finding volumes, not areas. Therefore, B is the most appropriate choice.

📖 Explanation: The curves intersect when $y^2 = 2 - y^2$, so $2y^2 = 2$, giving $y = \pm 1$. The region extends from $y = -1$ to $y = 1$. The right curve is $x = 2 - y^2$ and the left curve is $x = y^2$. The area is $\int_{-1}^{1} (2 - y^2 - y^2) \, dy = \int_{-1}^{1} (2 - 2y^2) \, dy = [2y - \frac{2y^3}{3}]_{-1}^{1} = (2 - \frac{2}{3}) - (-2 + \frac{2}{3}) = \frac{4}{3} - (-\frac{4}{3}) = \frac{8}{3}$. Option A is correct. Option B would be the area if the integrand were doubled incorrectly. Option C is half the correct area. Option D is the same as A but repeated.

📖 Explanation: The integral $\int_{-2}^{2} (4 - y^2) \, dy$ is correct for the area between $x = y^2$ and $x = 4$. Evaluating: $[4y - \frac{y^3}{3}]_{-2}^{2} = (8 - \frac{8}{3}) - (-8 + \frac{8}{3}) = \frac{16}{3} - (-\frac{16}{3}) = \frac{32}{3}$. The student's result is correct. Option A correctly states that the integral represents the area. The symmetry of the region means integrating from -2 to 2 gives the total area directly. Options B and C are incorrect because they misinterpret the setup. Option D is partially correct but the integral is correct regardless of symmetry considerations; the symmetry just makes it easier.

📖 Explanation: For $y \ge 0$, the curves intersect at $y = 0$ and $y = 1$. On $(0,1)$, $y > y^3$, so the right curve is $x = y$ and the left is $x = y^3$. The area is $\int_{0}^{1} (y - y^3) \, dy$. Evaluating: $[\frac{y^2}{2} - \frac{y^4}{4}]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$. Option A is correct. Option B reverses the integrand, which would give a negative area. Option C and D include negative y-values, which would give the total area including the symmetric region, but the question specifically asks for $y \ge 0$.

📖 Explanation: The curve $x = y^2 - 1$ intersects the y-axis when $x = 0$, so $y^2 - 1 = 0$, giving $y = \pm 1$. The region between the curve and the y-axis extends from $y = -1$ to $y = 1$. The right boundary is the y-axis ($x = 0$) and the left boundary is the curve ($x = y^2 - 1$), but note that $y^2 - 1 \le 0$ on $[-1,1]$. The length of the horizontal cross-section is $0 - (y^2 - 1) = 1 - y^2$. Thus the area is $\int_{-1}^{1} (1 - y^2) \, dy$. Option B is correct. Option A would give a negative area. Option C uses absolute value, which is equivalent but not simplified. Option D only integrates half the region.

📖 Explanation: The integral $\int_{0}^{1} (2y - y^2) \, dy$ suggests that the right boundary is $x = 2y$ and the left boundary is $x = y^2$ because $2y - y^2 > 0$ on $(0,1)$. These curves intersect when $2y = y^2$, so $y(y - 2) = 0$, giving $y = 0$ or $y = 2$, not $y = 1$. Wait, this is a discrepancy: if the integrand is $2y - y^2$, the intersection points are $y = 0$ and $y = 2$. But the limits are 0 to 1, which means the region is only part of the full region between the curves. This could be a region bounded by the curves and a horizontal line $y = 1$. Option A identifies the curves as $x = 2y$ and $x = y^2$, which matches the integrand. Option B reverses them. Option C is incorrect about the intersection points. Option D introduces a different curve.

📖 Explanation: The curves $x = y^2$ and $x = 2 - y^2$ intersect when $y^2 = 2 - y^2$, so $2y^2 = 2$, giving $y = \pm 1$. Substituting $y = 1$ into either equation gives $x = 1$. Substituting $y = -1$ gives $x = 1$. So the intersection points are $(1, 1)$ and $(1, -1)$. Option D correctly lists the points as $(1, 1)$ and $(1, -1)$ but the order is reversed; the points are $(1, -1)$ and $(1, 1)$. However, option D is the closest to correct. Option A has points with x and y swapped incorrectly. Option B has $(1, -1)$ and $(1, 1)$ which is correct but the order is not important. Option C has $(-1, -1)$ which is incorrect. So the correct answer is B or D depending on how the points are listed. Since D lists $(1,1)$ and $(-1,1)$, which are the correct points but the second point should be $(1,-1)$, the exact correct answer would be B if listed as $(1, -1)$ and $(1, 1)$.

📖 Explanation: The curves $x = \cos y$ and $x = \sin y$ intersect when $\cos y = \sin y$, which occurs at $y = \pi/4$ in the interval $[0, \pi/2]$. For $y < \pi/4$, $\cos y > \sin y$, so the right curve is $x = \cos y$. For $y > \pi/4$, $\sin y > \cos y$, so the right curve changes to $x = \sin y$. To compute the area correctly, the integral must be split at the intersection point to account for this change. Option B correctly explains the necessity of the split. Option C is true but doesn't explain why the split is needed. Option D is a consequence of the change. Option A is incorrect because the functions are not periodic in this interval.

📖 Explanation: The curves $x = y^2 - 4$ and $x = 4 - y^2$ intersect when $y^2 - 4 = 4 - y^2$, so $2y^2 = 8$, giving $y = \pm 2$. For $-2 < y < 2$, $4 - y^2 > y^2 - 4$, so the right curve is $x = 4 - y^2$ and the left is $x = y^2 - 4$. The integrand is $(4 - y^2) - (y^2 - 4) = 8 - 2y^2$. Thus the area is $\int_{-2}^{2} (8 - 2y^2) \, dy$. Option A is correct. Options B and D only use one of the curves. Option C uses the difference with the wrong sign and would give a negative area.

📖 Explanation: The curve $x = 3 - y^2$ intersects the y-axis when $x = 0$, so $3 - y^2 = 0$, giving $y = \pm \sqrt{3}$. The region between the curve and the y-axis extends from $y = -\sqrt{3}$ to $y = \sqrt{3}$. The student's limits 0 to 3 are incorrect: they used the x-intercept value (3) as a y-limit, and they only integrated from 0 to $\sqrt{3}$ for the upper half. The correct integral for the full region is $\int_{-\sqrt{3}}^{\sqrt{3}} (3 - y^2) \, dy$, or equivalently $2\int_{0}^{\sqrt{3}} (3 - y^2) \, dy$. Option B correctly identifies that the limits should be $-\sqrt{3}$ to $\sqrt{3}$. Option A is incorrect because the upper limit should be $\sqrt{3}$, not 3. Option C suggests the wrong sign. Option D suggests doubling the integral, which would be correct if the limits were 0 to $\sqrt{3}$, but the student's limits are 0 to 3, so doubling would still be wrong.

📖 Explanation: The formula for the area between two curves when integrating with respect to $y$ requires that the right curve $f(y)$ is always greater than or equal to the left curve $g(y)$ on the interval $[c,d]$. This ensures the integrand is nonnegative and represents the length of the horizontal cross-section. Option C states this condition. While continuity (option A) is needed for integrability, it's not sufficient; the inequality must hold. Option B is not required because the curves can cross the y-axis. Option D is not necessary; the region could be bounded by vertical lines or the y-axis itself.

📖 Explanation: The curves intersect when $y^2 = y + 2$, so $y^2 - y - 2 = 0$, giving $y = -1$ and $y = 2$. On $(-1, 2)$, $y + 2 \ge y^2$. The area is $\int_{-1}^{2} (y + 2 - y^2) \, dy$. Evaluating: $[ \frac{y^2}{2} + 2y - \frac{y^3}{3} ]_{-1}^{2} = (2 + 4 - \frac{8}{3}) - (\frac{1}{2} - 2 + \frac{1}{3}) = \frac{10}{3} - (-\frac{7}{6}) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$. Option A is correct. Option B is the area when integrating with respect to x for the same region (from Example 4 in the text). Option C and D are areas of subregions if split incorrectly.

📖 Explanation: Integrating with respect to $y$ is advantageous when the region has a simple description in terms of $y$—i.e., when the boundaries are easily expressed as functions of $y$ and the limits of integration in $y$ are easy to find. This often occurs when the curves are given as $x = f(y)$ and the region extends over a simple interval in y. Option B captures this idea. While option A is true (you need functions of y), it doesn't explain the advantage. Option C is not necessarily an advantage; integration with respect to y might still require splitting. Option D is not a sufficient condition; symmetry could also be handled with x-integration.

📖 Explanation: The integral $\int_{0}^{2} (2y - y^2) \, dy = [y^2 - \frac{y^3}{3}]_0^2 = 4 - \frac{8}{3} = \frac{12 - 8}{3} = \frac{4}{3}$. Option A is correct. Option B would be the area if the limits were -2 to 2 for a symmetric region. Option C is half the correct area. Option D is one-third of the correct area.

📖 Explanation: The curve $x = y^2 + 1$ intersects the y-axis when $y^2 + 1 = 0$, which has no real solution, so the region is bounded by the y-axis, the curve, and the horizontal line $y = 2$. However, the curve is always to the right of the y-axis ($x > 0$), so the horizontal cross-sections from the y-axis ($x = 0$) to the curve have length $y^2 + 1$. The limits of integration are from $y = 0$ to $y = 2$ if we consider the region in the first quadrant. But the curve is symmetric about the x-axis, so if the region is between the curve and the y-axis for $y$ from -2 to 2, the area would be $\int_{-2}^{2} (y^2 + 1) \, dy$. The question likely implies the region in the first quadrant, so option A is correct. Option B would be for the full symmetric region. Option C misses the constant term. Option D is the same as A.

📖 Explanation: For the region between $x = y^2$ and $x = 4$, integrating with respect to $y$ gives the area as $\int_{-2}^{2} (4 - y^2) \, dy$, a single integral. Integrating with respect to $x$ would require splitting the region into two parts because the left boundary is the parabola, which has two branches: $y = \sqrt{x}$ and $y = -\sqrt{x}$. The area would be $2\int_{0}^{4} \sqrt{x} \, dx$ if symmetric, which is still manageable, but the point is that integrating with respect to y avoids the need to consider the two branches. Option B correctly points out that the y-integral is a single integral. Option A is incorrect because $y = \pm \sqrt{x}$ is not easier to integrate in this context. Option C is subjective. Option D is not entirely true because you don't always need inverse functions if you split correctly.