Volumes with Cross Sections: Squares/Rectangles
Help Questions
AP Calculus AB › Volumes with Cross Sections: Squares/Rectangles
What is the correct volume integral for the solid with base between $y=x^2$ and $y=4$, square cross sections perpendicular to the $x$-axis?
$\displaystyle \int_{0}^{4} \big(\sqrt{y}\big)^2,dy$
$\displaystyle \int_{-2}^{2} \big(4-x^2\big)^2,dx$
$\displaystyle \int_{-2}^{2} \left(\frac{4-x^2}{2}\right)^2,dx$
$\displaystyle \int_{-2}^{2} \big(4+x^2\big)^2,dx$
$\displaystyle \int_{-2}^{2} \big(4-x^2\big),dx$
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.
What is the correct volume integral setup if the base is bounded by $y=\cos x$ and $y=0$ on $0,\tfrac{\pi}{2}$ with square cross sections perpendicular to the $x$-axis?
$\displaystyle \int_{0}^{\pi/2} (\sin x)^2,dx$
$\displaystyle \int_{0}^{\pi/2} \cos x,dx$
$\displaystyle \int_{0}^{\pi/2} (\cos x)^2,dx$
$\displaystyle \int_{0}^{1} (\arccos y)^2,dy$
$\displaystyle \int_{0}^{\pi/2} (\tfrac{\pi}{2}-\cos x)^2,dx$
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.
What is the correct volume integral setup if the base is bounded by $x=y+2$ and $x=0$ for $y\in-2,1$ with rectangular cross sections perpendicular to the $y$-axis of height $4$?
$\displaystyle \int_{-2}^{1} (y+2),dy$
$\displaystyle \int_{-2}^{1} 4\big(0-(y+2)\big),dy$
$\displaystyle \int_{-2}^{1} 4\big((y+2)-0\big)^2,dy$
$\displaystyle \int_{-2}^{1} 4\big((y+2)-0\big),dy$
$\displaystyle \int_{0}^{2} 4\big((y+2)-0\big),dy$
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.
What is the correct volume integral setup if the base is bounded by $y=2-x$ and $y=0$ on $0,2$ with square cross sections perpendicular to the $x$-axis?
$\displaystyle \int_{0}^{2} (x-2)^2,dx$
$\displaystyle \int_{0}^{2} (2-x),dx$
$\displaystyle \int_{0}^{2} (2-x^2)^2,dx$
$\displaystyle \int_{0}^{2} (2-\sqrt{x})^2,dx$
$\displaystyle \int_{0}^{2} (2-x)^2,dx$
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.
What is the correct volume integral setup if the base is bounded by $y=3$ and $y=\tfrac{x}{2}$ on $0,6$ with rectangular cross sections perpendicular to the $x$-axis of height $x$?
$\displaystyle \int_{0}^{6} \left(3-\frac{x}{2}\right),dx$
$\displaystyle \int_{0}^{3} x\left(3-\frac{x}{2}\right),dx$
$\displaystyle \int_{0}^{6} x\left(\frac{x}{2}-3\right),dx$
$\displaystyle \int_{0}^{6} x\left(3-\frac{x}{2}\right),dx$
$\displaystyle \int_{0}^{6} x\left(3-\frac{x}{2}\right)^2,dx$
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 = 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.
What is the correct volume integral setup if the base is bounded by $x=9-y^2$ and $x=0$ with square cross sections perpendicular to the $y$-axis?
$\displaystyle \int_{-3}^{3} (y^2-9)^2,dy$
$\displaystyle \int_{-3}^{3} (9-y^2)^2,dy$
$\displaystyle \int_{-3}^{3} (9-y^2),dy$
$\displaystyle \int_{0}^{3} (9-y^2)^2,dy$
$\displaystyle \int_{0}^{9} (9-x^2)^2,dx$
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 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.
What is the correct volume integral setup if the base is bounded by $x=\ln(y+1)$ and $x=0$ for $y\in0,e-1$ with square cross sections perpendicular to the $y$-axis?
$\displaystyle \int_{0}^{e-1} \ln(y+1),dy$
$\displaystyle \int_{0}^{e-1} (\ln y)^2,dy$
$\displaystyle \int_{0}^{1} (\ln(y+1))^2,dy$
$\displaystyle \int_{0}^{e-1} (\ln(y+1))^2,dy$
$\displaystyle \int_{1}^{e} (\ln y)^2,dy$
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.
What is the correct volume integral setup if the base is bounded by $x=\sqrt{y}$ and $x=0$ for $y\in0,4$ with square cross sections perpendicular to the $y$-axis?
$\displaystyle \int_{0}^{4} (y)^2,dy$
$\displaystyle \int_{0}^{4} (\sqrt{y})^2,dy$
$\displaystyle \int_{0}^{4} (4-\sqrt{y})^2,dy$
$\displaystyle \int_{0}^{2} (\sqrt{y})^2,dy$
$\displaystyle \int_{0}^{4} \sqrt{y},dy$
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.
What is the correct volume integral setup if the base is bounded by $y=2\sin x$ and $y=0$ on $0,\pi$ with square cross sections perpendicular to the $x$-axis?
$\displaystyle \int_{0}^{\pi} 2\sin x,dx$
$\displaystyle \int_{0}^{\pi} (2\cos x)^2,dx$
$\displaystyle \int_{0}^{\pi} (2\sin x)^2,dx$
$\displaystyle \int_{0}^{\pi} (\sin x)^2,dx$
$\displaystyle \int_{0}^{2} (2\sin x)^2,dx$
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.
What is the correct volume integral setup if the base is bounded by $x=4$ and $x=1+y$ for $y\in0,3$ with rectangular cross sections perpendicular to the $y$-axis of height $y$?
$\displaystyle \int_{0}^{3} (4-(1+y)),dy$
$\displaystyle \int_{0}^{3} y\big(4-(1+y)\big),dy$
$\displaystyle \int_{1}^{4} y\big(4-(1+y)\big),dy$
$\displaystyle \int_{0}^{3} y\big((1+y)-4\big),dy$
$\displaystyle \int_{0}^{3} y\big(4-(1+y)\big)^2,dy$
Explanation
This problem requires using the method of cross-sectional volumes to find the volume of a solid where the cross-sections are 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.