AP Calculus BC Quiz: Volumes With Cross Sections Squares Rectangles

What this quiz covers

This quiz focuses on Volumes With Cross Sections Squares Rectangles, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus BC.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} (2-y^2)^2\,dy$ (correct answer)
2. $\displaystyle \int_{0}^{2} (2-y^2)^2\,dy$
3. $\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} (2-y^2)\,dy$
4. $\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} (2+y^2)^2\,dy$
5. $\displaystyle \int_{-2}^{2} (2-y^2)^2\,dy$

Explanation: This problem requires finding volume with square cross-sections perpendicular to the y-axis. The region is bounded by x = 2 and x = y² for -√2 ≤ y ≤ √2. These boundaries intersect where y² = 2, giving y = ±√2. For any y-value in this interval, the horizontal distance is 2 - y². Since cross-sections are squares, the side length equals 2 - y², making the cross-sectional area (2-y²)². Choice C gives only the side length without squaring, which would not provide the correct area for square cross-sections. The systematic method is to determine the width of the base region at each y-value, square this distance to obtain the area of each square cross-section, then integrate along the y-axis.

Question 2

1. $\displaystyle \int_{1}^{e} 2(1-\ln x)\,dx$ (correct answer)
2. $\displaystyle \int_{1}^{e} 2(\ln x-1)\,dx$
3. $\displaystyle \int_{1}^{e} (1-\ln x)^2\,dx$
4. $\displaystyle \int_{1}^{e} 2(\ln x)^2\,dx$
5. $\displaystyle \int_{0}^{1} 2(1-\ln x)\,dy$

Explanation: This problem requires computing volume with rectangular cross-sections perpendicular to the x-axis, where the base width comes from the region and height is fixed at 2. The base is bounded by y = ln x and y = 1 for 1 ≤ x ≤ e. For any x-value, we need to determine which function is on top: at x = e, ln e = 1, so they meet at the right endpoint. For 1 ≤ x ≤ e, we have ln x ≤ 1, so the width is 1 - ln x. The cross-sectional area is width × height = (1 - ln x) × 2 = 2(1 - ln x). Choice C incorrectly squares the width, treating cross-sections as squares rather than rectangles with height 2. The key strategy for rectangular cross-sections is to identify which boundary is upper, find the width, multiply by the specified height, then integrate.

Question 3

1. $\displaystyle \int_{0}^{\pi} (\sin x)^2\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{\pi} \sin x\,dx$
3. $\displaystyle \int_{0}^{\pi} (\sin x)^3\,dx$
4. $\displaystyle \int_{0}^{1} (\arcsin y)^2\,dy$
5. $\displaystyle \int_{0}^{\pi} (\pi-\sin x)^2\,dx$

Explanation: This problem tests cross-sectional volume reasoning for solids with square cross-sections. The base region is bounded by y = sin x and y = 0 from x = 0 to π. The cross-sections are perpendicular to the x-axis, so the side length of each square is sin x. The area of each cross-section is (sin x)². A tempting distractor is choice B, which is ∫ sin x dx, but that computes the area of the base rather than the volume. A transferable strategy for cross-section volumes is to identify the integration variable, express the side length(s) in terms of that variable based on the base boundaries, compute the cross-sectional area, and integrate over the interval.

Question 4

1. $\displaystyle \int_{1}^{9} (\sqrt{x})^2\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{3} (y^2)^2\,dy$
3. $\displaystyle \int_{1}^{9} \sqrt{x}\,dx$
4. $\displaystyle \int_{1}^{9} (x)^2\,dx$
5. $\displaystyle \int_{1}^{9} (9-x)^2\,dx$

Explanation: This problem tests cross-sectional volume reasoning for solids with square cross-sections. The base region is bounded by x = 1, x = 9, y = 0, and y = √x from x = 1 to x = 9. The cross-sections are perpendicular to the x-axis, so the side length of each square is √x. The area of each cross-section is (√x)² = x. A tempting distractor is choice C, which is ∫ √x dx, but that computes the area of the base rather than the volume. A transferable strategy for cross-section volumes is to identify the integration variable, express the side length(s) in terms of that variable based on the base boundaries, compute the cross-sectional area, and integrate over the interval.

Question 5

1. $\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} 5(2-x^2)\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{2} 5(2-y^2)\,dy$
3. $\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} 5(2-x^2)^2\,dx$
4. $\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} (2-x^2)\,dx$
5. $\displaystyle \int_{-\sqrt{2}}^{\sqrt{2}} (5-x^2)(2-x^2)\,dx$

Explanation: This problem tests cross-sectional volume reasoning for solids with rectangular cross-sections. The base region is between y = x² and y = 2 from x = -√2 to √2. The cross-sections are perpendicular to the x-axis, so the base length of each rectangle is 2 - x². The height is a constant 5, so the area is 5(2 - x²). A tempting distractor is choice D, which omits the height factor of 5 and computes something closer to the base area. A transferable strategy for cross-section volumes is to identify the integration variable, express the side length(s) in terms of that variable based on the base boundaries, compute the cross-sectional area, and integrate over the interval.

Question 6

1. $\displaystyle \int_{1}^{e} \ln(x)\,dx$
2. $\displaystyle \int_{0}^{1} \ln(x)^2\,dx$
3. $\displaystyle \int_{1}^{e} \big(\ln(x)\big)^2\,dx$ (correct answer)
4. $\displaystyle \int_{1}^{e} \big(\ln(x)\big)^3\,dx$
5. $\displaystyle \int_{1}^{e} (e-x)^2\,dx$

Explanation: This problem asks for the volume when square cross-sections are perpendicular to the x-axis. The base is bounded by y = ln(x) and y = 0 for 1 ≤ x ≤ e, so the side length of each square equals ln(x) - 0 = ln(x). Since we have square cross-sections, the area is (side length)² = (ln(x))². The integral is ∫_{1}^{e} (ln(x))² dx. Choice A incorrectly uses just ln(x) without squaring, which would give the area under the curve rather than the volume. For square cross-sections, always square the distance between boundary curves to obtain the cross-sectional area.

Question 7

1. $\displaystyle \int_{0}^{6} (6-x)^2\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{6} (6-x)\,dx$
3. $\displaystyle \int_{0}^{6} (6-x)^3\,dx$
4. $\displaystyle \int_{0}^{6} (6)^2\,dx$
5. $\displaystyle \int_{0}^{6} x^2\,dx$

Explanation: This problem asks for the volume when the base is bounded by y = 6 - x and y = 0, with square cross-sections perpendicular to the x-axis. At each x-value between 0 and 6, the side length of the square equals the vertical distance (6 - x) - 0 = 6 - x. Since we have square cross-sections, the area of each square is (side length)² = (6 - x)². The integral becomes ∫_{0}^{6} (6 - x)² dx. Choice B incorrectly uses just the linear expression without squaring it, which would give the area under the curve rather than the volume. Remember that for square cross-sections, you must square the distance between boundaries to find the cross-sectional area.

Question 8

1. $\displaystyle \int_{0}^{2}\big(2x-x^2\big)^2\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{2}\big(x^2-2x\big)^2\,dx$
3. $\displaystyle \int_{0}^{2}\big(2x-x^2\big)\,dx$
4. $\displaystyle \int_{0}^{2}\big(2- x\big)^2\,dx$
5. $\displaystyle \int_{0}^{2}\big(2x-x^2\big)^2\,dy$

Explanation: This problem involves finding the volume of a solid with square cross-sections perpendicular to the x-axis. The base region is bounded by y = 2x and y = x², which intersect when 2x = x², giving x = 0 and x = 2. For each x-value between 0 and 2, since 2x ≥ x² in this interval, the side length of the square equals 2x - x². The area of each square cross-section is (side length)² = (2x - x²)². Choice C incorrectly uses just the side length without squaring it, which would give the area of the base region rather than the volume. Remember that for square cross-sections, you must square the expression for the side length before integrating.

Question 9

1. $\displaystyle \int_{0}^{2} (2x-x^2)^2\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{2} (2x-x^2)\,dx$
3. $\displaystyle \int_{0}^{2} (x^2-2x)^2\,dx$
4. $\displaystyle \int_{0}^{2} \big(2x^2-x\big)^2\,dx$
5. $\displaystyle \int_{0}^{2} (2x-x^2)^2\,dy$

Explanation: This problem involves finding the volume when square cross sections are perpendicular to the x-axis. The base region is bounded by y = 2x (upper curve) and y = x² (lower curve) for 0 ≤ x ≤ 2. At each x-value, the vertical distance between curves is (2x - x²), which becomes the side length of each square cross section. The area of each square is therefore (2x - x²)², and integrating this from 0 to 2 gives the volume. Choice B shows just (2x - x²) without squaring, which would give the area of the base region, not the volume. Remember that for square cross sections, you must square the distance between boundary curves to get the cross-sectional area.

Question 10

1. $\displaystyle \int_{0}^{2}\big(4-y^2\big)^2\,dy$ (correct answer)
2. $\displaystyle \int_{0}^{2}\big(4-y^2\big)\,dy$
3. $\displaystyle \int_{0}^{4}\big(4-y^2\big)^2\,dy$
4. $\displaystyle \int_{0}^{2}\big(4-y^2\big)^2\,dx$
5. $\displaystyle \int_{0}^{2}\big(2-\sqrt{y}\big)^2\,dy$

Explanation: This problem requires finding the volume of a solid with square cross-sections perpendicular to the y-axis. The base region lies between x = y² and x = 4 for 0 ≤ y ≤ 2, so for each y-value, the side length of the square equals the horizontal distance: 4 - y². Since we need square cross-sections, the area of each square is (side length)² = (4 - y²)². The integration variable must be y since we're slicing perpendicular to the y-axis, and the limits are from y = 0 to y = 2. Choice B incorrectly uses just the side length without squaring it, while choice D incorrectly uses dx instead of dy. When slicing perpendicular to the y-axis, always integrate with respect to y and square the side length for square cross-sections.

Question 11

1. $\displaystyle \int_{-3}^{3}(9-x^2)\,dx$
2. $\displaystyle \int_{-3}^{3}(9-x^2)^2\,dx$ (correct answer)
3. $\displaystyle \int_{0}^{3}(9-x^2)^2\,dx$
4. $\displaystyle \int_{-3}^{3}(3-x)^2\,dx$
5. $\displaystyle \int_{-3}^{3}(9-x^2)^2\,dy$

Explanation: This problem requires finding the volume of a solid with square cross-sections perpendicular to the x-axis. The base region is bounded by the parabola y = 9 - x² and y = 0, which intersect when 9 - x² = 0, giving x = ±3. For each x-value, the side length of the square equals the vertical distance 9 - x². Since we need square cross-sections, the area of each square is (side length)² = (9 - x²)². We integrate from x = -3 to x = 3 to cover the entire base region. Choice A incorrectly uses just the side length without squaring it, which would give the area under the curve rather than the volume of the solid. Always square the side length expression when dealing with square cross-sections.

Question 12

1. $\displaystyle \int_{0}^{1} 2(3x)^2\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{1} (3x)\cdot 2\,dx$
3. $\displaystyle \int_{0}^{1} (3x)^2\,dx$
4. $\displaystyle \int_{0}^{3} 2y^2\,dy$
5. $\displaystyle \int_{0}^{1} \pi\,2(3x)^2\,dx$

Explanation: This problem requires cross-sectional volume reasoning to compute the volume with rectangular cross sections perpendicular to the x-axis. The base region is bounded by x = 0, x = 1, y = 0, and y = 3x. For each x, the base of the rectangle is the vertical length 3x, and the height is twice that, so 6x. Therefore, the area is 3x * 6x = 18x², or equivalently 2*(3x)², and the integral is from 0 to 1. A tempting distractor is choice E, which includes π, but that's for circular shapes, not rectangles. In general, for volumes with cross sections perpendicular to an axis, identify the varying dimension as the side length or radius depending on the shape, square it for area, and integrate along the axis.

Question 13

1. $\displaystyle \int_{0}^{2} x(2-x)\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{2} x(2-x)^2\,dx$
3. $\displaystyle \int_{0}^{2} (2-x)\,dx$
4. $\displaystyle \int_{0}^{2} (x-2)(2-x)\,dx$
5. $\displaystyle \int_{0}^{2} x(2-y)\,dy$

Explanation: This problem requires computing volume with rectangular cross-sections perpendicular to the x-axis, where the base width comes from the region and height varies as x. The base is bounded by y = 2 - x and y = 0 on [0,2]. For any x-value, the width is (2-x) - 0 = 2-x. The cross-sectional area is width × height = (2-x) × x = x(2-x). Choice B incorrectly squares the width, treating cross-sections as squares rather than rectangles with height x. The systematic approach for rectangular cross-sections is to identify the base width from the region, multiply by the specified variable height, then integrate this area function over the given interval.

Question 14

1. $\displaystyle \int_{0}^{2} (4-y^2)^2\,dy$ (correct answer)
2. $\displaystyle \int_{0}^{4} (4-\sqrt{x})^2\,dx$
3. $\displaystyle \int_{0}^{2} (4-y^2)\,dy$
4. $\displaystyle \int_{-2}^{2} (4-y^2)^2\,dx$
5. $\displaystyle \int_{0}^{2} \pi(4-y^2)^2\,dy$

Explanation: This problem involves cross-sectional volume reasoning to determine the volume of a solid with square cross sections perpendicular to the y-axis. The base region is bounded by x = y² and x = 4, considering y from 0 to 2 in the first quadrant. For each y, the side length of the square is the horizontal distance between x = y² and x = 4, which is 4 - y². Therefore, the cross-sectional area is (4 - y²)², and the volume is the integral from 0 to 2. A tempting distractor is choice E, which includes π, but that's incorrect for square cross sections as it suggests a circular shape. In general, for volumes with cross sections perpendicular to an axis, identify the varying dimension as the side length or radius depending on the shape, square it for area, and integrate along the axis.

Question 15

1. $\displaystyle \int_{0}^{1} (x-x^2)^2\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{1} (x-x^2)\,dx$
3. $\displaystyle \int_{0}^{1} \big(\sqrt{x}-x\big)^2\,dx$
4. $\displaystyle \int_{0}^{1} (x-x^2)^2\,dy$
5. $\displaystyle \int_{0}^{1} \pi(x-x^2)^2\,dx$

Explanation: This problem involves cross-sectional volume reasoning to compute the volume of a solid with square cross sections perpendicular to the x-axis. The base region is bounded by the curves y = x and y = x² from x = 0 to x = 1. At each x in this interval, the side length of the square cross section is the vertical distance between the curves, which is x - x² since y = x is above y = x². Therefore, the area of each cross section is (x - x²)², and the volume is obtained by integrating this area function from 0 to 1. A tempting distractor is choice E, which includes a factor of π, but this is incorrect because π is used for circular cross sections, not squares. In general, when finding volumes with known cross sections, determine the area function based on the shape and integrate it over the base interval.

Question 16

1. $\displaystyle \int_{0}^{2} (2x-x^2)^2\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{2} (x^2-2x)^2\,dx$
3. $\displaystyle \int_{0}^{2} (2x-x^2)\,dx$
4. $\displaystyle \int_{0}^{2} \big(\sqrt{y}-\tfrac{y}{2}\big)^2\,dy$
5. $\displaystyle \int_{0}^{2} \pi(2x-x^2)^2\,dx$

Explanation: This problem requires cross-sectional volume reasoning to compute the volume with square cross sections perpendicular to the x-axis. The base region is bounded by y = 2x and y = x² from x = 0 to x = 2. At each x, the side length is the vertical distance between y = 2x (upper) and y = x² (lower), which is 2x - x². Thus, the area is (2x - x²)², and the volume integral is from 0 to 2. A tempting distractor is choice E, which adds π, but this is for circular cross sections, not squares. In general, when finding volumes with known cross sections, determine the area function based on the shape and integrate it over the base interval.

Question 17

1. $\displaystyle \int_{-2}^{2} (4-x^2)^2\,dx$ (correct answer)
2. $\displaystyle \int_{-2}^{2} (4-x^2)\,dx$
3. $\displaystyle \int_{0}^{4} \big(\sqrt{4-y}-(-\sqrt{4-y})\big)^2\,dy$
4. $\displaystyle \int_{-2}^{2} \pi(4-x^2)^2\,dx$
5. $\displaystyle \int_{-2}^{2} (2\sqrt{4-x^2})^2\,dx$

Explanation: This problem requires cross-sectional volume reasoning to find the volume of a solid with square cross sections perpendicular to the x-axis. The base region is bounded by y = 4 - x² and y = 0 from x = -2 to x = 2. For each x, the side length of the square is the height of the parabola above the x-axis, which is 4 - x². Thus, the cross-sectional area is (4 - x²)², and the volume is the integral of this from -2 to 2. A tempting distractor is choice D, which adds π, but that's for disks or washers, not square cross sections. In general, for volumes with cross sections perpendicular to an axis, identify the varying dimension as the side length or radius depending on the shape, square it for area, and integrate along the axis.

Question 18

1. $\displaystyle \int_{0}^{1} (1-y)^2\,dy$ (correct answer)
2. $\displaystyle \int_{0}^{1} (1-y)\,dy$
3. $\displaystyle \int_{0}^{1} (1+y)^2\,dy$
4. $\displaystyle \int_{0}^{1} (1-x)^2\,dx$
5. $\displaystyle \int_{0}^{1} \left(\frac{1-y}{2}\right)^2\,dy$

Explanation: This problem involves computing volume with square cross-sections perpendicular to the y-axis. The base region is bounded by x = 0 and x = 1 - y for 0 ≤ y ≤ 1. For any y-value, the horizontal distance between these boundaries is (1 - y) - 0 = 1 - y. Since cross-sections are squares, this distance serves as the side length, so the area is (1 - y)². Choice B represents only the side length rather than the area of the square. The key method is to find the width of the base region at each y-value, use this as the square's side length, then integrate the squared expression along the y-axis.

Question 19

1. $\displaystyle \int_{0}^{\pi} (2-\sin x)^2\,dx$ (correct answer)
2. $\displaystyle \int_{0}^{\pi} (2-\sin x)\,dx$
3. $\displaystyle \int_{0}^{\pi} (2+\sin x)^2\,dx$
4. $\displaystyle \int_{0}^{2} (2-\sin x)^2\,dy$
5. $\displaystyle \int_{0}^{\pi} \left(\frac{2-\sin x}{2}\right)^2\,dx$

Explanation: This problem requires computing volume using square cross-sections perpendicular to the x-axis. The base region is bounded by y = sin x and y = 2 on the interval [0,π]. For any x-value, the vertical distance between these curves is 2 - sin x. Since the cross-sections are squares, this distance serves as the side length, making the area of each square (2 - sin x)². Choice B represents only the side length rather than the area of the square cross-section. The correct method is to find the distance between bounding curves, use it as the square's side length, then integrate the squared expression over the given interval.

Question 20

1. $\displaystyle \int_{0}^{1} 2(1-y)\,dy$ (correct answer)
2. $\displaystyle \int_{0}^{1} 2(1-y)^2\,dy$
3. $\displaystyle \int_{0}^{1} (1-y)\,dy$
4. $\displaystyle \int_{0}^{1} \frac{1-y}{2}\,dy$
5. $\displaystyle \int_{0}^{1} 2(1-x)\,dx$

Explanation: This problem involves finding volume with rectangular cross-sections perpendicular to the y-axis, where one dimension is the width of the base region and the other is a fixed height of 2. The base is bounded by x = 1 - y and x = 0 for 0 ≤ y ≤ 1. For any y-value, the width is (1-y) - 0 = 1-y. The cross-sectional area is width × height = (1-y) × 2 = 2(1-y). Choice B incorrectly squares the width, treating it as if cross-sections were squares rather than rectangles. The correct approach for rectangular cross-sections is to multiply the base width by the given height, then integrate this area function over the appropriate interval.