AP Calculus AB 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 AB.

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

Explanation: This problem requires finding the volume of a solid using cross-sections perpendicular to the x-axis. The base region lies between y = x² and y = 4, so for each x-value, the height of the base is (4 - x²). Since the cross-sections are squares perpendicular to the x-axis, each square has side length equal to this height, giving an area of (4 - x²)². The x-values range from -2 to 2 (where the curves intersect), so the volume integral is ∫_{-2}^{2} (4 - x²)² dx. Choice A incorrectly uses just (4 - x²) without squaring, which would give the area of the base region rather than the volume. When working with square cross-sections, always square the expression for the side length to get the cross-sectional area.

Question 2

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

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid with square cross sections. The base of the solid is the region bounded by y = cos x and y = 0 on [0, π/2]. For cross sections perpendicular to the x-axis, the side length of each square is the vertical distance between the curves, which is cos x. Therefore, the area of each cross section is (cos x)². Integrating this area function from 0 to π/2 gives the volume as ∫ from 0 to π/2 of (cos x)² dx. A common mistake is choice B, which forgets to square the side length. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

Question 3

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

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid with rectangular cross sections. The base of the solid is the region bounded by x = y + 2 and x = 0 for y in [-2, 1]. For cross sections perpendicular to the y-axis, the width of each rectangle is the horizontal distance between the curves, which is y + 2. The height of each rectangle is given as 4, so the area is 4(y + 2). Integrating this area function from -2 to 1 gives the volume as ∫ from -2 to 1 of 4((y + 2) - 0) dy. A tempting distractor is choice B, which squares the width incorrectly. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

Question 4

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

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid with square cross sections. The base of the solid is the region bounded by y = 2 - x and y = 0 on [0, 2]. For cross sections perpendicular to the x-axis, the side length of each square is the vertical distance between the curves, which is 2 - x. Therefore, the area of each cross section is (2 - x)². Integrating this area function from 0 to 2 gives the volume as ∫ from 0 to 2 of (2 - x)² dx. A common mistake is choice B, which forgets to square the side length. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

Question 5

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

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid with rectangular cross sections. The base of the solid is the region bounded by y = 3 and y = x/2 on [0, 6]. For cross sections perpendicular to the x-axis, the width of each rectangle is the vertical distance between the curves, which is 3 - x/2. The height of each rectangle is given as x, so the area is x(3 - x/2). Integrating this area function from 0 to 6 gives the volume as ∫ from 0 to 6 of x(3 - x/2) dx. A common mistake is choice B, which squares the width unnecessarily. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

Question 6

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

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid with square cross sections. The base of the solid is the region bounded by x = 9 - y² and x = 0. For cross sections perpendicular to the y-axis from y = -3 to y = 3, the side length of each square is the horizontal distance between the curves, which is 9 - y². Therefore, the area of each cross section is (9 - y²)². Integrating this area function from -3 to 3 gives the volume as ∫ from -3 to 3 of (9 - y²)² dy. A tempting distractor is choice C, which omits the squaring and computes incorrectly. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

Question 7

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

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross-sections are squares perpendicular to the y-axis. The base of the solid is the region bounded by x = ln(y + 1) and x = 0 from y = 0 to y = e - 1, so at each y, the side length of the square is ln(y + 1). Since the cross-sections are squares, the area of each cross-section is (ln(y + 1))^2. The volume is obtained by integrating this area function from y = 0 to y = e - 1. A tempting distractor is choice B, which integrates ln(y + 1) dy without squaring, but that computes the area of the base rather than the volume. Remember, for volumes with known cross-sections, always determine the area of the cross-section as a function of the variable of integration and integrate that over the interval.

Question 8

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

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross-sections are squares perpendicular to the y-axis. The base of the solid is the region bounded by x = √y and x = 0 from y = 0 to y = 4, so at each y, the side length of the square is √y. Since the cross-sections are squares, the area of each cross-section is (√y)^2. The volume is obtained by integrating this area function from y = 0 to y = 4. A tempting distractor is choice B, which integrates √y dy without squaring, but that computes the area of the base rather than the volume. Remember, for volumes with known cross-sections, always determine the area of the cross-section as a function of the variable of integration and integrate that over the interval.

Question 9

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

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross-sections are squares perpendicular to the x-axis. The base of the solid is the region bounded by y = 2 sin x and y = 0 from x = 0 to x = π, so at each x, the side length of the square is 2 sin x. Since the cross-sections are squares, the area of each cross-section is (2 sin x)^2. The volume is obtained by integrating this area function from x = 0 to x = π. A tempting distractor is choice B, which integrates 2 sin x dx, but that computes the area of the base rather than the volume with square cross-sections. Remember, for volumes with known cross-sections, always determine the area of the cross-section as a function of the variable of integration and integrate that over the interval.

Question 10

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

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross-sections are rectangles perpendicular to the y-axis. The base of the solid is the region bounded by x = 4 and x = 1 + y from y = 0 to y = 3, so at each y, the width of the rectangle is 4 - (1 + y). The height of each rectangle is given as y, so the area is y (4 - (1 + y)). The volume is obtained by integrating this area function from y = 0 to y = 3. A tempting distractor is choice D, which integrates (4 - (1 + y)) dy, but that omits the rectangular height and gives the base area. Remember, for volumes with known cross-sections, always determine the area of the cross-section as a function of the variable of integration and integrate that over the interval.

Question 11

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

Explanation: This problem involves finding the volume when square cross-sections are perpendicular to the x-axis. The base region is bounded by y = sin x (upper curve) and y = 0 (lower curve, the x-axis) on the interval [0, π]. At each x-value, the square's side length equals the vertical distance between curves: sin x - 0 = sin x. Since a square with side length s has area s², the cross-sectional area is (sin x)² = sin² x. Choice A incorrectly uses sin x without squaring, which would represent a rectangle of height 1, not a square. When working with cross-sections, always square the side length for squares or use the appropriate area formula for the given shape.

Question 12

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

Explanation: This problem involves finding the volume of a solid with square cross-sections perpendicular to the y-axis. The region is bounded by x = y and x = 2 - y over the interval [0, 1], so at each y-value, the side length of the square is the horizontal distance between the curves: (2 - y) - y = 2 - 2y. Since the cross-sections are squares, the area of each square is (2 - 2y)². The volume is obtained by integrating this area from y = 0 to y = 1, yielding ∫₀¹ (2 - 2y)² dy. Choice A incorrectly uses (2 - y)² instead of (2 - 2y)², failing to account for the full distance between the curves. For cross-sectional volume problems, carefully identify which variable you're integrating with respect to and ensure your distance calculation matches that orientation.

Question 13

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

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid with rectangular cross sections. The base of the solid is the region bounded by y = ln x and y = 0 on [1, e]. For cross sections perpendicular to the x-axis, the width of each rectangle is the vertical distance between the curves, which is ln x. The height of each rectangle is given as 5, so the area is 5 ln x. Integrating this area function from 1 to e gives the volume as ∫ from 1 to e of 5 ln x dx. A common mistake is choice B, which squares the width as if for squares. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

Question 14

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

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid with square cross sections. The base of the solid is the region bounded by y = x + 1 and y = 0 on [-1, 2]. For cross sections perpendicular to the x-axis, the side length of each square is the vertical distance between the curves, which is x + 1. Therefore, the area of each cross section is (x + 1)². Integrating this area function from -1 to 2 gives the volume as ∫ from -1 to 2 of (x + 1)² dx. A tempting distractor is choice B, which omits the squaring and computes area instead of volume. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

Question 15

1. $\displaystyle \int_{-2}^{2} 6\big((1+y^2)-1\big)\,dy$ (correct answer)
2. $\displaystyle \int_{-2}^{2} 6\big(1-(1+y^2)\big)\,dy$
3. $\displaystyle \int_{-2}^{2} 6\big((1+y^2)-1\big)^2\,dy$
4. $\displaystyle \int_{0}^{4} 6(y^2)\,dy$
5. $\displaystyle \int_{-2}^{2} (1+y^2)\,dy$

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid with rectangular cross sections. The base of the solid is the region bounded by x = 1 and x = 1 + y² for y in [-2, 2]. For cross sections perpendicular to the y-axis, the width of each rectangle is the horizontal distance between the curves, which is (1 + y²) - 1 = y². The height of each rectangle is given as 6, so the area is 6 y². Integrating this area function from -2 to 2 gives the volume as ∫ from -2 to 2 of 6((1 + y²) - 1) dy. A common mistake is choice C, which squares the width incorrectly for rectangles. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

Question 16

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. $\displaystyle \int_{0}^{2} (2-x)x^2\,dx$
5. $\displaystyle \int_{0}^{2} (x-2)x\,dx$

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid with rectangular cross sections. The base of the solid is the region bounded by y = 2 and y = x on [0, 2]. For cross sections perpendicular to the x-axis, the width of each rectangle is the vertical distance between the curves, which is 2 - x. The height of each rectangle is given as x, so the area is x(2 - x). Integrating this area function from 0 to 2 gives the volume as ∫ from 0 to 2 of x(2 - x) dx. A common mistake is choice B, which squares the width unnecessarily. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

Question 17

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

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid with square cross sections. The base of the solid is the region bounded by y = 2x and y = x². For cross sections perpendicular to the x-axis from x = 0 to x = 2, the side length of each square is the vertical distance between the curves, which is 2x - x². Therefore, the area of each cross section is (2x - x²)². Integrating this area function from 0 to 2 gives the volume as ∫ from 0 to 2 of (2x - x²)² dx. A common mistake is choice B, which forgets to square the side length and integrates the distance directly. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

Question 18

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

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid with rectangular cross sections. The base of the solid is the region bounded by y = 4 - x² and y = 0 on [-2, 2]. For cross sections perpendicular to the x-axis, the width of each rectangle is the vertical distance between the curves, which is 4 - x². The height of each rectangle is given as 3, so the area is 3(4 - x²). Integrating this area function from -2 to 2 gives the volume as ∫ from -2 to 2 of 3(4 - x²) dx. A tempting distractor is choice B, which squares the width incorrectly for rectangles. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

Question 19

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

Explanation: This problem involves finding the volume of a solid with rectangular cross-sections perpendicular to the y-axis. The base region lies between x = 2y and x = y², where these curves intersect at y = 0 and y = 2. For 0 ≤ y ≤ 2, we have 2y ≥ y², so the base of each rectangle is (2y - y²). With constant height 3, each rectangle has area 3(2y - y²). The volume integral is ∫_{0}^{2} 3(2y - y²) dy. Choice C incorrectly uses (2y - y²)² without the height factor, treating the cross-sections as squares instead of rectangles with height 3. For rectangular cross-sections with given height, multiply the base width by the specified height.

Question 20

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

Explanation: This problem involves finding the volume of a solid with square cross-sections perpendicular to the y-axis. The base region lies between x = 0 and x = ln(y + 1) for 0 ≤ y ≤ e - 1. For each y-value, the width of the base is ln(y + 1) - 0 = ln(y + 1). Since the cross-sections are squares, each has side length ln(y + 1) and area [ln(y + 1)]². The volume integral is ∫_{0}^{e-1} [ln(y + 1)]² dy. Choice A incorrectly uses ln(y + 1) without squaring, which would give only the area of the base region. When working with cross-sections perpendicular to the y-axis, always express the integrand in terms of y and square the side length for square cross-sections.