Cross Sections: Squares & Rectangles - Calculus 1

The volume of a solid is represented by the definite integral $V = \int_1^e (\ln x)^2 \,dx$ . Which of the following descriptions matches this solid?
Answer: The solid with a base bounded by $y=\ln x$ , $y=0$ , $x=1$ , and $x=e$ , with square cross-sections perpendicular to the x-axis. The integral for the volume of a solid with known cross-sections is $V = \int_a^b A(x) \,dx$ for cross-sections perpendicular to the x-axis. Here, the integral is with respect to $x$ from $1$ to $e$ . The integrand is $A(x) = (\ln x)^2$ . Since this represents the area of a square cross-section, $A(x) = [s(x)]^2$ , the side length must be $s(x) = \ln x$ . This side length represents the height of the region from the x-axis ( $y=0$ ) up to the curve $y=\ln x$ . Thus, the solid has a base bounded by $y=\ln x$ , $y=0$ , from $x=1$ to $x=e$ , with square cross-sections perpendicular to the x-axis.
A solid's base is the region in the first quadrant bounded by $x+y=4$ . Cross-sections perpendicular to the x-axis are rectangles whose height is equal to their base. What is the volume?
Answer: $\frac{64}{3}$ Rectangles whose height is equal to their base are squares. The base is a triangle with vertices (0,0), (4,0), and (0,4). We integrate along the x-axis from 0 to 4. The bounding line is $y = 4-x$ . The side length of a square cross-section is $s(x) = (4-x) - 0 = 4-x$ . The area is $A(x) = (4-x)^2 = 16 - 8x + x^2$ . The volume is $V = \int_0^4 (16 - 8x + x^2) \,dx = [16x - 4x^2 + \frac{x^3}{3}]_0^4 = 16(4) - 4(16) + \frac{64}{3} = 64 - 64 + \frac{64}{3} = \frac{64}{3}$ .
The base of a solid is the region bounded by $y = \ln(x)$ , $y=0$ , and $x=e^2$ . Cross-sections taken perpendicular to the x-axis are rectangles whose height is a constant $h=3$ . Which integral represents the volume of this solid?
Answer: $\int_1^{e^2} 3 \ln x \,dx$ The region is bounded by $y=\ln x$ and $y=0$ . The intersection with $y=0$ is at $\ln x = 0 \implies x=1$ . The limits of integration are from $x=1$ to $x=e^2$ . The base of a rectangular cross-section is the height of the region, $s(x) = \ln x$ . The height of the rectangle is given as a constant, 3. The area of the cross-section is $A(x) = \text{base} \times \text{height} = (\ln x) \cdot 3 = 3 \ln x$ . The volume is the integral of this area, $V = \int_1^{e^2} 3 \ln x \,dx$ .
A solid has a base in the xy-plane defined by the region enclosed by $y=4$ and $y=x^2$ . Cross-sections perpendicular to the y-axis are squares. Which expression gives the volume of this solid?
Answer: $\int_{0}^{4} 4y \,dy$ Since cross-sections are perpendicular to the y-axis, we integrate with respect to $y$ . The region extends from $y=0$ (the vertex of the parabola) to $y=4$ . We express $x$ in terms of $y$ : $x = \pm \sqrt{y}$ . At a given height $y$ , the horizontal chord across the region extends from $x = -\sqrt{y}$ to $x = \sqrt{y}$ . The length of this chord, which is the side of the square, is $s(y) = \sqrt{y} - (-\sqrt{y}) = 2\sqrt{y}$ . The area of the square cross-section is $A(y) = [s(y)]^2 = (2\sqrt{y})^2 = 4y$ . The volume is the integral of this area from $y=0$ to $y=4$ , which is $\int_{0}^{4} 4y \,dy$ .
Let $R$ be the region enclosed by the graph of $y = \frac{1}{x}$ , the x-axis, and the lines $x=1$ and $x=5$ . A solid is formed with base $R$ and cross-sections perpendicular to the x-axis that are rectangles of constant height 3. What is the volume of the solid?
Answer: $3 \ln(5)$ The integration is with respect to $x$ from $1$ to $5$ . For a given $x$ , the base of the rectangular cross-section is the height of the region $R$ , which is $s(x) = \frac{1}{x} - 0 = \frac{1}{x}$ . The height of the rectangle is given as a constant, $h=3$ . The area of a cross-section is $A(x) = s(x) \cdot h = \frac{1}{x} \cdot 3 = \frac{3}{x}$ . The volume is $V = \int_1^5 \frac{3}{x} \,dx = 3[\ln|x|]_1^5 = 3(\ln(5) - \ln(1)) = 3\ln(5)$ .
The base of a solid is the circle $x^2 + y^2 = 9$ . The cross-sections perpendicular to the x-axis are squares. What is the volume of the solid?
Answer: 144 The limits of integration along the x-axis are from $x=-3$ to $x=3$ . For a given $x$ , the vertical chord of the circle has endpoints on $y = \sqrt{9-x^2}$ and $y = -\sqrt{9-x^2}$ . The length of this chord, which is the side length of the square cross-section, is $s(x) = \sqrt{9-x^2} - (-\sqrt{9-x^2}) = 2\sqrt{9-x^2}$ . The area of the square is $A(x) = [s(x)]^2 = (2\sqrt{9-x^2})^2 = 4(9-x^2) = 36 - 4x^2$ . The volume is $V = \int_{-3}^3 (36 - 4x^2) \,dx$ . Since the integrand is even, $V = 2 \int_0^3 (36 - 4x^2) \,dx = 2[36x - \frac{4}{3}x^3]_0^3 = 2(36(3) - \frac{4}{3}(27)) = 2(108 - 36) = 2(72) = 144$ .
Let a solid have a base defined by the region between $y = \sqrt{x}$ and the x-axis from $x=0$ to $x=4$ . Cross-sections perpendicular to the x-axis are rectangles whose height is half of their base. What is the volume of the solid?
Answer: 4 The limits of integration are given as $x=0$ to $x=4$ . The base of a rectangular cross-section at a given $x$ is $s(x) = \sqrt{x} - 0 = \sqrt{x}$ . The height is half the base, so $h(x) = \frac{1}{2}s(x) = \frac{1}{2}\sqrt{x}$ . The area of the cross-section is $A(x) = s(x) \cdot h(x) = \sqrt{x} \cdot \frac{1}{2}\sqrt{x} = \frac{1}{2}x$ . The volume is the integral of the area: $V = \int_0^4 \frac{1}{2}x \,dx = \frac{1}{2}[\frac{x^2}{2}]_0^4 = \frac{1}{4}[x^2]_0^4 = \frac{1}{4}(4^2 - 0^2) = \frac{16}{4} = 4$ .
The base of a solid is the region in the first quadrant enclosed by the graphs of $y = \sqrt{x}$ and $y = x^2$ . The cross-sections of the solid perpendicular to the x-axis are squares. What is the volume of the solid?
Answer: $9/70$ First, find the points of intersection by setting the functions equal: $\sqrt{x} = x^2 \implies x = x^4 \implies x^4 - x = 0 \implies x(x^3 - 1) = 0$ . The intersection points are at $x=0$ and $x=1$ . In the interval $(0, 1)$ , $\sqrt{x} > x^2$ . The side length $s(x)$ of a square cross-section at $x$ is the vertical distance between the curves, $s(x) = \sqrt{x} - x^2$ . The area of the square cross-section is $A(x) = s(x)^2 = (\sqrt{x} - x^2)^2 = x - 2x^{5/2} + x^4$ . The volume is the integral of the area from $x=0$ to $x=1$ : $V = \int_{0}^{1} (x - 2x^{5/2} + x^4) \,dx = \left[ \frac{1}{2}x^2 - 2\left(\frac{2}{7}x^{7/2}\right) + \frac{1}{5}x^5 \right]_{0}^{1} = \frac{1}{2} - \frac{4}{7} + \frac{1}{5} = \frac{35 - 40 + 14}{70} = \frac{9}{70}$ .
Let $R$ be a region in the xy-plane that forms the base of two different solids. Solid $S_1$ has cross-sections perpendicular to the x-axis that are squares. Solid $S_2$ has cross-sections perpendicular to the x-axis that are rectangles with height three times their base. If the volume of $S_1$ is 15, what is the volume of $S_2$ ?
Answer: $45$ When you encounter problems involving solids with known cross-sections, you're working with the method of cross-sectional areas. The key insight is that both solids share the same base region $R$ , so at any x-value, their cross-sections have the same width. Let's say the width of the region at position $x$ is $w(x)$ . For solid $S_1$ , each cross-section is a square with side length $w(x)$ , giving area $[w(x)]^2$ . For solid $S_2$ , each cross-section is a rectangle where the base is $w(x)$ and the height is $3w(x)$ (three times the base), giving area $w(x) \cdot 3w(x) = 3[w(x)]^2$ . Notice that at every x-value, the area of $S_2$ 's cross-section is exactly 3 times the area of $S_1$ 's cross-section. When you integrate to find volumes: $V_{S_1} = \int [w(x)]^2 dx = 15$ and $V_{S_2} = \int 3[w(x)]^2 dx = 3\int [w(x)]^2 dx = 3 \cdot 15 = 45$ The answer is B) 45. Choice A) 135 likely comes from incorrectly multiplying by 9 (perhaps squaring the factor of 3). Choice C) 5 suggests dividing by 3 instead of multiplying. Choice D) $15\sqrt{3}$ might result from confusion about how the geometric relationship translates to volume. Strategy tip: In cross-section problems, always identify the relationship between the areas (not just the linear dimensions) of corresponding cross-sections. The volume ratio equals the area ratio since both solids extend over the same interval.
The base of a solid is the region bounded by $y = x^3$ , the line $x = 2$ , and the x-axis. The cross-sections perpendicular to the y-axis are rectangles with a height equal to one-fourth of their base. What is the volume of the solid?
Answer: $4/5$ This problem tests your ability to set up volume integrals when cross-sections are perpendicular to the y-axis rather than the x-axis. When you see "perpendicular to the y-axis," you need to think of thin horizontal slices and integrate with respect to y. First, identify the region. The solid's base is bounded by $y = x^3$ , $x = 2$ , and the x-axis, so it extends from $x = 0$ to $x = 2$ and from $y = 0$ to $y = 8$ (since when $x = 2$ , $y = 2^3 = 8$ ). For integration with respect to y, express x in terms of y: from $y = x^3$ , we get $x = y^{1/3}$ . At height y, the rectangular cross-section has a base extending from $x = 0$ to $x = y^{1/3}$ , so the base length is $y^{1/3}$ . The height is one-fourth of the base, which is $\frac{1}{4}y^{1/3}$ . The area of each rectangular cross-section is: $A(y) = y^{1/3} \cdot \frac{1}{4}y^{1/3} = \frac{1}{4}y^{2/3}$ The volume is: $V = \int_0^8 \frac{1}{4}y^{2/3} dy = \frac{1}{4} \cdot \frac{3}{5}y^{5/3}\Big|_0^8 = \frac{3}{20}(8^{5/3}) = \frac{3}{20}(32) = \frac{4}{5}$ Answer D is correct. Answer A (24/5) likely comes from incorrect integration bounds or forgetting the 1/4 factor. Answer B (16/5) suggests integrating $y^{2/3}$ without the 1/4 coefficient. Answer C (32/7) indicates using the wrong power in integration. Strategy tip: When cross-sections are perpendicular to the y-axis, always solve for x in terms of y and integrate with respect to y.
The base of a solid is the region enclosed by $x=y^2$ and $x=4$ . Cross-sections perpendicular to the y-axis are rectangles whose height is equal to the y-coordinate of the cross-section. What is the volume of the solid?
Answer: 0 The region is bounded by $x=y^2$ and $x=4$ . The intersection points are where $y^2=4$ , so $y=\pm 2$ . We integrate with respect to $y$ from -2 to 2. The base of a rectangular cross-section is $s(y) = 4 - y^2$ . The height is given as $h(y)=y$ . The area of a cross-section is $A(y) = s(y) \cdot h(y) = (4-y^2)y = 4y - y^3$ . The volume is $V = \int_{-2}^2 (4y - y^3) \,dy$ . Since the integrand $f(y) = 4y - y^3$ is an odd function ( $f(-y) = -f(y)$ ) and the interval of integration is symmetric about the origin, the value of the definite integral is 0.
The base of a solid is the region enclosed by $y = \sec(x)$ , the x-axis, $x = -\pi/4$ , and $x = \pi/4$ . Cross-sections perpendicular to the x-axis are squares. What is the volume of the solid?
Answer: 2 The limits of integration are from $x = -\pi/4$ to $x = \pi/4$ . The side length of a square cross-section is $s(x) = \sec(x) - 0 = \sec(x)$ . The area of the cross-section is $A(x) = [s(x)]^2 = \sec^2(x)$ . The volume is the integral of the area: $V = \int_{-\pi/4}^{\pi/4} \sec^2(x) \,dx$ . The antiderivative of $\sec^2(x)$ is $\tan(x)$ . So, $V = [\tan(x)]_{-\pi/4}^{\pi/4} = \tan(\pi/4) - \tan(-\pi/4) = 1 - (-1) = 2$ .
A solid is constructed on a base bounded by the curves $y = e^x$ , $y=1$ , and $x=2$ . The cross-sections perpendicular to the x-axis are squares. The volume of the solid is:
Answer: $\frac{1}{2}e^4 - 2e^2 + \frac{7}{2}$ The region is bounded by $y=e^x$ (top curve) and $y=1$ (bottom curve). The intersection occurs where $e^x = 1$ , which is $x=0$ . The region extends from $x=0$ to $x=2$ . The side length of a square cross-section is $s(x) = e^x - 1$ . The area of a cross-section is $A(x) = (e^x - 1)^2 = e^{2x} - 2e^x + 1$ . The volume is $V = \int_0^2 (e^{2x} - 2e^x + 1) \,dx = [\frac{1}{2}e^{2x} - 2e^x + x]_0^2 = (\frac{1}{2}e^4 - 2e^2 + 2) - (\frac{1}{2}e^0 - 2e^0 + 0) = \frac{1}{2}e^4 - 2e^2 + 2 - (\frac{1}{2} - 2) = \frac{1}{2}e^4 - 2e^2 + 2 + \frac{3}{2} = \frac{1}{2}e^4 - 2e^2 + \frac{7}{2}$ .
The base of a solid is the region enclosed by the graph of $x = 4 - y^2$ and the y-axis. If the cross-sections perpendicular to the y-axis are squares, what is the volume of the solid?
Answer: $512/15$ The region is bounded by the parabola $x = 4 - y^2$ and the line $x=0$ . The parabola intersects the y-axis at $y = \pm 2$ . Since the cross-sections are perpendicular to the y-axis, we integrate with respect to $y$ . The side length $s(y)$ of a square cross-section is the horizontal distance from the y-axis to the curve, so $s(y) = 4 - y^2$ . The area is $A(y) = s(y)^2 = (4 - y^2)^2 = 16 - 8y^2 + y^4$ . The volume is $V = \int_{-2}^{2} (16 - 8y^2 + y^4) \,dy$ . Since the integrand is an even function, we can simplify: $V = 2 \int_{0}^{2} (16 - 8y^2 + y^4) \,dy = 2 \left[ 16y - \frac{8}{3}y^3 + \frac{1}{5}y^5 \right]_{0}^{2} = 2 \left( 32 - \frac{64}{3} + \frac{32}{5} \right) = 2 \left( \frac{480 - 320 + 96}{15} \right) = 2 \left( \frac{256}{15} \right) = \frac{512}{15}$ .
The region bounded by $y = x^2 + 1$ , $y=0$ , $x=0$ , and $x=2$ is the base of a solid. Cross-sections perpendicular to the x-axis are squares. Find the volume.
Answer: $\frac{206}{15}$ This problem tests your understanding of volumes using cross-sections, a key application of integration in Calculus 1. When you see "cross-sections perpendicular to an axis," think about slicing the solid and finding the area of each slice. First, you need to understand the base region. It's bounded by $y = x^2 + 1$ (top), $y = 0$ (bottom), $x = 0$ (left), and $x = 2$ (right). At any x-value, the height of this region is $(x^2 + 1) - 0 = x^2 + 1$ . Since cross-sections perpendicular to the x-axis are squares, and the side length of each square equals the height of the base region, each square has side length $x^2 + 1$ . The area of each square cross-section is therefore $(x^2 + 1)^2$ . To find the volume, integrate these cross-sectional areas from $x = 0$ to $x = 2$ : $V = \int_0^2 (x^2 + 1)^2 \, dx = \int_0^2 (x^4 + 2x^2 + 1) \, dx$ $= \left[\frac{x^5}{5} + \frac{2x^3}{3} + x\right]_0^2 = \frac{32}{5} + \frac{16}{3} + 2 = \frac{96 + 80 + 30}{15} = \frac{206}{15}$ Answer choice A ( $\frac{14}{3}$ ) likely comes from integrating just $x^2 + 1$ instead of squaring it. Choice B ( $\frac{56}{3}$ ) might result from incorrect expansion or integration errors. Choice C ( $\frac{32}{5}$ ) represents only the $\frac{x^5}{5}$ term, missing the other components. Key strategy: Always square the side length when dealing with square cross-sections, and carefully expand $(x^2 + 1)^2$ before integrating.
A solid has a base bounded by $y=\sin(x)$ and $y=\cos(x)$ between $x=0$ and $x=\pi/4$ . If cross-sections perpendicular to the x-axis are squares, what is the volume?
Answer: $\frac{\pi}{4} - \frac{1}{2}$ When you encounter a volume problem with cross-sections, you need to set up an integral where each slice has a known area. Here, the cross-sections are squares perpendicular to the x-axis, so you need to find the side length of each square at any x-value. First, determine which function is on top in the given interval. At $x = 0$ : $\sin(0) = 0$ and $\cos(0) = 1$ , so $\cos(x) > \sin(x)$ initially. The functions intersect when $\sin(x) = \cos(x)$ , which occurs at $x = \pi/4$ (where both equal $\sqrt{2}/2$ ). Throughout $[0, \pi/4]$ , cosine remains above sine. The side length of each square cross-section is the distance between the curves: $\cos(x) - \sin(x)$ . Since the area of a square is side length squared, each cross-section has area $[\cos(x) - \sin(x)]^2$ . The volume is: $V = \int_0^{\pi/4} [\cos(x) - \sin(x)]^2 \, dx$ Expanding: $[\cos(x) - \sin(x)]^2 = \cos^2(x) - 2\sin(x)\cos(x) + \sin^2(x) = 1 - 2\sin(x)\cos(x) = 1 - \sin(2x)$ Therefore: $V = \int_0^{\pi/4} [1 - \sin(2x)] \, dx = [x + \frac{1}{2}\cos(2x)]_0^{\pi/4} = [\frac{\pi}{4} + 0] - [0 + \frac{1}{2}] = \frac{\pi}{4} - \frac{1}{2}$ Answer D is correct. Options A and B likely come from integration errors or incorrect setup. Option C represents $\cos(\pi/4) - \sin(0) = \sqrt{2}/2 - 0$ , suggesting confusion about what to integrate. Strategy tip: Always expand squared expressions completely and use trigonometric identities to simplify before integrating.
The base of a solid is the region in the first quadrant bounded by the graph of $y = 4 - x^2$ and the coordinate axes. Cross-sections perpendicular to the x-axis are squares. What is the volume of the solid?
Answer: $\frac{256}{15}$ The region is bounded by $y=4-x^2$ , $y=0$ , and $x=0$ . The x-intercept of the parabola is at $x=2$ , so the integration is from $x=0$ to $x=2$ . The side length of a square cross-section at a given $x$ is $s(x) = (4-x^2) - 0 = 4-x^2$ . The area of the cross-section is $A(x) = [s(x)]^2 = (4-x^2)^2 = 16 - 8x^2 + x^4$ . The volume is the integral of the area: $V = \int_0^2 (16 - 8x^2 + x^4) \,dx = [16x - \frac{8}{3}x^3 + \frac{1}{5}x^5]_0^2 = (16(2) - \frac{8}{3}(8) + \frac{1}{5}(32)) - 0 = 32 - \frac{64}{3} + \frac{32}{5} = \frac{480 - 320 + 96}{15} = \frac{256}{15}$ .
The base of a solid is a triangle in the xy-plane with vertices at $(0,0)$ , $(4,0)$ , and $(0,2)$ . The cross-sections of the solid perpendicular to the x-axis are squares. What is the volume of the solid?
Answer: $16/3$ This problem involves finding the volume of a solid using cross-sections, a key technique in Calculus 1. When you see "cross-sections perpendicular to an axis," you're integrating the area of those cross-sections along that axis. First, you need to understand the triangular base. The vertices $(0,0)$ , $(4,0)$ , and $(0,2)$ form a right triangle. The hypotenuse connects $(4,0)$ to $(0,2)$ , with equation $y = 2 - \frac{x}{2}$ . Since cross-sections perpendicular to the x-axis are squares, at any x-value between 0 and 4, the square's side length equals the height of the triangle at that x-coordinate. From the line equation, this height is $2 - \frac{x}{2}$ . The area of each square cross-section is therefore $A(x) = \left(2 - \frac{x}{2}\right)^2 = 4 - 2x + \frac{x^2}{4}$ . To find the volume, integrate from $x = 0$ to $x = 4$ : $V = \int_0^4 \left(4 - 2x + \frac{x^2}{4}\right) dx = \left[4x - x^2 + \frac{x^3}{12}\right]_0^4$ $V = 16 - 16 + \frac{64}{12} = \frac{16}{3}$ Answer choice A ( $8$ ) likely comes from using just the side length times the base. Choice B ( $32/3$ ) might result from incorrectly squaring the integral bounds. Choice C ( $4$ ) could come from confusing this with a simpler geometric calculation. Remember: for cross-section problems, always identify the shape's dimensions at each point, square or cube as needed for area/volume, then integrate along the specified axis.
The base of a solid is the region enclosed by the graphs of $y = \sec(x)$ , $y = \tan(x)$ , $x = 0$ , and $x = \pi/4$ . Cross-sections perpendicular to the x-axis are squares. Which integral represents the volume of the solid?
Answer: $\int_{0}^{\pi/4} (\sec(x) - \tan(x))^2 \,dx$ When finding volumes of solids with known cross-sections, you need to identify the area of each cross-section as a function of x, then integrate along the axis of cross-sections. First, visualize the region. Between $x = 0$ and $x = \pi/4$ , $\sec(x)$ lies above $\tan(x)$ (since $\sec^2(x) - \tan^2(x) = 1 > 0$ , so $\sec(x) > \tan(x)$ in this interval where both are positive). The region is bounded by these curves and the vertical lines $x = 0$ and $x = \pi/4$ . For any value of x in this interval, the cross-section is a square whose side length equals the vertical distance between the curves: $\sec(x) - \tan(x)$ . Since the area of a square equals (side length)², the cross-sectional area is $(\sec(x) - \tan(x))^2$ . The volume is therefore $\int_{0}^{\pi/4} (\sec(x) - \tan(x))^2 \,dx$ , which is answer D. Answer A gives $\int_{0}^{\pi/4} \sec^2(x) \,dx$ , which would be correct if the side length were $\sec(x)$ instead of the difference between the functions. Answer B gives $\int_{0}^{\pi/4} (\sec^2(x) - \tan^2(x)) \,dx = \int_{0}^{\pi/4} 1 \,dx$ , which treats the integrand as if it were already squared. Answer C incorrectly includes the factor $\pi$ , suggesting confusion with circular cross-sections rather than squares. Remember: for volume problems with known cross-sections, the key steps are (1) find the dimension(s) of each cross-section as functions of the integration variable, (2) calculate the area using the appropriate geometric formula, then (3) integrate.
Let $R$ be the region enclosed by the graphs of $y = x^2$ and $y = 4$ . A solid has $R$ as its base. For this solid, each cross-section perpendicular to the x-axis is a rectangle whose height is twice the length of its base. Find the volume of the solid.
Answer: $1024/15$ The region $R$ is bounded by $y=4$ (top) and $y=x^2$ (bottom). The intersection points are at $x^2 = 4$ , so $x = \pm 2$ . The cross-sections are perpendicular to the x-axis. At a given $x$ , the base of the rectangle is the vertical distance $b(x) = 4 - x^2$ . The height is $h(x) = 2 \cdot b(x) = 2(4 - x^2)$ . The area of the cross-section is $A(x) = b(x)h(x) = (4 - x^2) \cdot 2(4 - x^2) = 2(4 - x^2)^2 = 2(16 - 8x^2 + x^4)$ . The volume is $V = \int_{-2}^{2} 2(16 - 8x^2 + x^4) \,dx = 2 \cdot 2 \int_{0}^{2} (16 - 8x^2 + x^4) \,dx = 4 \left[ 16x - \frac{8}{3}x^3 + \frac{1}{5}x^5 \right]_{0}^{2} = 4 \left( 32 - \frac{64}{3} + \frac{32}{5} \right) = 4 \left( \frac{256}{15} \right) = \frac{1024}{15}$ .

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