Free math worksheets

- The area function is already provided, so no base-width conversion is needed. - Integrating square units over inches gives cubic inches.

1. Volume accumulates cross-sectional area: \(V=\int_0^3A(x)\,\mathrm{d}x\). 2. \(V=\int_0^3(5+2x-x^2)\,\mathrm{d}x=15\,\text{in}^3\).

\(15\,\text{in}^3\)

- Read the radius expression from the representative cross section. - Use that radius in the circle-area formula. - Accumulate the circular areas from \(x=0\) to \(x=2\).

1. The diagram gives the radius \(r(x)=2-x\). 2. The cross-sectional area is \(A(x)=\pi(2-x)^2\). 3. Therefore, \(V=\pi\int_0^2(2-x)^2\,\mathrm{d}x=\frac{8\pi}{3}\).

\(\frac{8\pi}{3}\) cubic units

- Read the vertical distance across the base at the marked x-value. - That distance is a side length, not yet an area. - How do you convert a square's side length into its area?

1. At \(x=0.5\), the displayed vertical segment runs from \(y=0\) to \(y=1.5\), so its length is \(\frac{3}{2}\) units. 2. That segment is the side of the square cross section, so the area is \(\left(\frac{3}{2}\right)^2=\frac{9}{4}\) square units.

Side length: \(\frac{3}{2}\) units Cross-sectional area: \(\frac{9}{4}\) square units

- The labeled segment gives the full diameter, not the radius. - How are diameter and radius related? - A semicircle uses half of the corresponding circle's area.

1. The displayed diameter is \(6\,\text{cm}\), so the radius is \(3\,\text{cm}\). 2. A semicircle has half the area of a circle, so \(A=\frac{1}{2}\pi(3)^2=\frac{9\pi}{2}\,\text{cm}^2\).

\(\frac{9\pi}{2}\,\text{cm}^2\)

- Use the vertical width of the triangular base as the short diagonal. - Substitute that variable diagonal into the supplied hexagon-area formula. - Integrate from \(x=0\) to the x-intercept of \(y=2-x\).

1. The short diagonal is the vertical base width \(d(x)=2-x\). 2. The cross-sectional area is \(A(x)=\frac{\sqrt3}{2}(2-x)^2\). 3. Therefore, \(V=\frac{\sqrt3}{2}\int_0^2(2-x)^2\,\mathrm{d}x=\frac{4\sqrt3}{3}\).

\(\frac{4\sqrt3}{3}\) cubic units

- Compare the two cross-sectional areas for the same width. - The ratio between the areas is constant at every slice. - A constant cross-sectional area ratio gives the same volume ratio.

1. At each slice, \(A_S=w^2\). 2. At the same slice, \(A_C=\frac{\pi}{4}w^2\). 3. Therefore, \(A_C=\frac{\pi}{4}A_S\) at every position. 4. Integrating gives \(V_C=\frac{\pi}{4}V_S\). Since \(\frac{\pi}{4}<1\), Solid S has the larger volume.

\(V_C=\frac{\pi}{4}V_S\); Solid S has the larger volume.

- What two-dimensional quantity must be accumulated to obtain volume? - How is the area of a square related to its side length \(w(x)\)? - After correcting the area function, integrate it over the given slicing interval.

1. The expression \(4w(x)\) is the square's perimeter, not its area. 2. The correct cross-sectional area is \(A(x)=w(x)^2=4\sin^2x\). 3. Therefore, \(V=\int_0^\pi4\sin^2x\,\mathrm{d}x=2\pi\).

Andrei integrated perimeter. The correct volume is \(2\pi\) cubic units.

- Use the two distinct triangle dimensions exactly as given. - Only the base varies with position.

1. The cross-sectional area is \(A(x)=\frac12(\sqrt x)(5)=\frac52\sqrt x\). 2. \(V=\int_0^9\frac52\sqrt x\,\mathrm{d}x=45\).

\(45\) cubic units

- Use \(w(x)\) and \(2w(x)\) as the two rectangle dimensions. - Multiply the dimensions before integrating. - Use symmetry or a power-reduction identity to evaluate the cosine-squared integral.

1. Each cross-sectional area is \(A(x)=w(x)\cdot2w(x)=2\cos^2x\). 2. Therefore, \(V=\int_{-\pi/2}^{\pi/2}2\cos^2x\,\mathrm{d}x=\pi\).

\(\pi\) cubic units

- Square the entire width function. - Set up the exact integral before using numerical technology.

1. The cross-sectional area is \(A(x)=e^{-2x^2}\). 2. \(V=\int_0^2e^{-2x^2}\,\mathrm{d}x\). 3. Numerical evaluation gives \(V\approx0.627\) cubic units.

\(\approx0.627\) cubic units

- Compare shape area formulas using the same side length. - Carry the constant ratio from slices to the entire solid.

1. Square area is \(w(x)^2\). 2. Equilateral-triangle area is \(\frac{\sqrt3}{4}w(x)^2\). 3. The same constant ratio holds after integration. 4. The triangle-cross-section volume is \(\frac{\sqrt3}{4}(40)=10\sqrt3\).

\(10\sqrt3\) cubic units

- Use the two equal but perpendicular triangle dimensions. - The one-half factor remains part of every cross-sectional area.

1. The area is \(A(x)=\frac12(3-x)(3-x)=\frac12(3-x)^2\). 2. \(V=\frac12\int_0^3(3-x)^2\,\mathrm{d}x=\frac{9}{2}\).

\(\frac{9}{2}\) cubic units

- Factor each shape's area into a constant times the squared width. - Use the supplied integral once for each possible constant.

1. The shape factors multiplying \(w^2\) are \(1\), \(\frac{\pi}{4}\), \(\frac{\pi}{8}\), and \(\frac{\sqrt3}{4}\), respectively. 2. Multiplying by \(\int_a^b w(x)^2\,\mathrm{d}x=12\), option C gives \(12\cdot\frac{\pi}{8}=\frac{3\pi}{2}\). 3. Therefore, semicircles with diameter \(w\) meet the design requirement.

Choice C: semicircles with diameter \(w\).

- Read the vertical distance between the two displayed curves. - That distance is the side length, not the area, of each equilateral triangle. - Apply the equilateral-triangle area formula before integrating.

1. On \(0\le x\le2\), the vertical width is \(s(x)=2x-x^2\). 2. An equilateral triangle with side \(s\) has area \(\frac{\sqrt3}{4}s^2\), so \(A(x)=\frac{\sqrt3}{4}(2x-x^2)^2\). 3. The volume setup is \(V=\int_0^2\frac{\sqrt3}{4}(2x-x^2)^2\,\mathrm{d}x\).

\(A(x)=\frac{\sqrt3}{4}(2x-x^2)^2\) \(V=\int_0^2\frac{\sqrt3}{4}(2x-x^2)^2\,\mathrm{d}x\)

- Use the vertical distance between the two base curves as the square's side length. - Square that distance to obtain cross-sectional area. - Integrate the cross-sectional area over \(0\le x\le4\).

1. The side length of a square is the vertical width of the base: \(s(x)=\sqrt x-\frac{x}{2}\). 2. The cross-sectional area is \(A(x)=s(x)^2\). 3. Therefore, \(V=\int_0^4\left(\sqrt x-\frac{x}{2}\right)^2\,\mathrm{d}x=\frac{8}{15}\).

\(\frac{8}{15}\) cubic units

- Find the x-interval from the intersections of the base curves. - Use \(3-x^2\) as the rectangle's base dimension. - Multiply the two rectangle dimensions before integrating.

1. The base width is \(b(x)=3-x^2\), and the curves intersect at \(x=\pm\sqrt3\). 2. The rectangle height is \(3b(x)\), so its area is \(A(x)=b(x)\cdot3b(x)=3(3-x^2)^2\). 3. Therefore, \(V=\int_{-\sqrt3}^{\sqrt3}3(3-x^2)^2\,\mathrm{d}x=\frac{144\sqrt3}{5}\).

\(\frac{144\sqrt3}{5}\) cubic units

- Use the vertical separation of the base curves as the triangle's side length. - Apply the equilateral-triangle area formula to that variable side. - Integrate from the two intersection x-values.

1. The curves intersect at \(x=0\) and \(x=1\), and the triangle side length is \(s(x)=x-x^2\). 2. The cross-sectional area is \(A(x)=\frac{\sqrt3}{4}s(x)^2\). 3. Therefore, \(V=\frac{\sqrt3}{4}\int_0^1(x-x^2)^2\,\mathrm{d}x=\frac{\sqrt3}{120}\).

\(\frac{\sqrt3}{120}\) cubic units

- Use the vertical base width as the semicircle's diameter, not its radius. - Express semicircle area in terms of the diameter. - Integrate over the full interval from \(x=-2\) to \(x=2\).

1. The diameter is \(d(x)=4-x^2\) for \(-2\le x\le2\). 2. A semicircle with diameter \(d\) has area \(A=\frac{\pi}{8}d^2\). 3. Therefore, \(V=\frac{\pi}{8}\int_{-2}^{2}(4-x^2)^2\,\mathrm{d}x=\frac{64\pi}{15}\).

\(\frac{64\pi}{15}\) cubic units

- Convert the long diagonal into the side length of the regular hexagon. - The cross-sectional area depends on the square of the base width. - Use symmetry when evaluating the volume integral.

1. If the long diagonal is \(w(x)\), then the hexagon side length is \(s(x)=\frac{w(x)}{2}\). 2. Using \(A=\frac{3\sqrt3}{2}s^2\), the cross-sectional area is \(A(x)=\frac{3\sqrt3}{8}(1-x^2)^2\). 3. The volume is \(V=\frac{3\sqrt3}{8}\int_{-1}^{1}(1-x^2)^2\,\mathrm{d}x\). 4. Since \(\int_{-1}^{1}(1-x^2)^2\,\mathrm{d}x=\frac{16}{15}\), \(V=\frac{2\sqrt3}{5}\).

a) \(A(x)=\frac{3\sqrt3}{8}(1-x^2)^2\) b) \(\frac{2\sqrt3}{5}\) cubic units

- Cross sections perpendicular to the y-axis use horizontal base widths. - Subtract \(x=y^2\) from \(x=2y\) to obtain the square's side. - Square the side length before integrating with respect to \(y\).

1. The curves intersect at \(y=0\) and \(y=2\). 2. The square side is the horizontal width \(s(y)=2y-y^2\). 3. Therefore, \(V=\int_0^2(2y-y^2)^2\,\mathrm{d}y=\frac{16}{15}\).

\(\frac{16}{15}\) cubic units

- Use the horizontal distance from \(x=y^2\) to \(x=4\) as one rectangle dimension. - Use the separately stated height \(y+3\) as the other dimension. - Integrate the product over \(-2\le y\le2\).

1. The rectangle base is \(b(y)=4-y^2\), and its other dimension is \(y+3\). 2. Thus, the cross-sectional area is \(A(y)=(4-y^2)(y+3)\). 3. Therefore, \(V=\int_{-2}^{2}(4-y^2)(y+3)\,\mathrm{d}y=32\).

\(32\) cubic units

- A side length is a distance, so use an absolute value when the upper line changes. - Squaring an absolute value gives the same result as squaring its expression. - Locate the line intersection to explain the change in geometric order.

1. The vertical distance between the lines is \(s(x)=|(3-x)-x|=|3-2x|\). 2. A square cross section has area \(A(x)=s(x)^2=|3-2x|^2=(3-2x)^2\). 3. Squaring removes the sign, so one area formula works across the intersection at \(x=\frac{3}{2}\). 4. Therefore, \(V=\int_0^3(3-2x)^2\,\mathrm{d}x=9\).

\(9\) cubic units

- Use two grid points on each line to determine its equation. - Subtract the lower y-value from the upper y-value to get the square side. - Square the side length and integrate to the intersection.

1. From the graph, the upper boundary is \(y=3-\frac{1}{2}x\) and the lower boundary is \(y=\frac{1}{2}x\). Thus, the square side length is \(s(x)=3-x\). 2. The boundaries meet at \(x=3\). 3. Therefore, \(V=\int_0^3(3-x)^2\,\mathrm{d}x=9\).

\(9\) cubic units

- The square side length is the given vertical width. - Squaring the width makes the volume depend on \(k^2\). - Use \(k>0\) after solving the volume equation.

1. The square area is \(A(x)=k^2x^2(1-x)^2\). 2. Therefore, \(V=k^2\int_0^1x^2(1-x)^2\,\mathrm{d}x=\frac{k^2}{30}\). 3. Set \(\frac{k^2}{30}=\frac{6}{5}\). Then \(k^2=36\), and \(k=6\) because \(k>0\).

\(k=6\)

- Express the volume in terms of the unknown endpoint. - Cancel the common geometric constant before solving.

1. The cross-sectional area is \(A(x)=\frac{\pi}{8}x^2\). 2. \(V=\frac{\pi}{8}\int_0^b x^2\,\mathrm{d}x=\frac{\pi b^3}{24}\). 3. Set \(\frac{\pi b^3}{24}=9\pi\), giving \(b^3=216\) and \(b=6\).

\(b=6\)

- What is the vertical distance between the two base boundaries at position \(x\)? - Compute the model volume before applying the physical scale factor. - How does volume scale when every linear dimension is multiplied by \(2\)?

1. In coordinate units, \(V_m=\int_0^4\left(2-\frac{x}{2}\right)^2\,\mathrm{d}x=\frac{16}{3}\). 2. A linear scale factor of \(2\) multiplies volume by \(2^3=8\). 3. Therefore, the actual volume is \(8V_m=\frac{128}{3}\,\text{cm}^3\).

\(\frac{128}{3}\,\text{cm}^3\)

- How do the two base curves determine the vertical width at a given \(x\)? - Express both triangle legs in terms of that width before finding area. - Volume is the integral of the resulting cross-sectional area.

1. The base width is \(w(x)=1-x^2\) for \(-1\le x\le1\). 2. The triangle area is \(A(x)=\frac{1}{2}w(x)[2w(x)]=w(x)^2\). 3. Therefore, \(V=\int_{-1}^{1}(1-x^2)^2\,\mathrm{d}x=\frac{16}{15}\).

\(\frac{16}{15}\) cubic units

- The vertical base width is the semicircle's diameter, not its radius. - Write semicircle area in terms of diameter before integrating. - What is \(\int_0^\pi\sin^2x\,\mathrm{d}x\)?

1. The diameter is \(d(x)=\sin x\). 2. The cross-sectional area is \(A(x)=\frac{\pi}{8}\sin^2x\). 3. Therefore, \(V=\frac{\pi}{8}\int_0^\pi\sin^2x\,\mathrm{d}x=\frac{\pi^2}{16}\).

\(\frac{\pi^2}{16}\) cubic units

- Use the base width as the square's diagonal, not its side. - Express square area in terms of diagonal length. - Integrate over the intersections \(x=-2\) and \(x=2\).

1. The square diagonal is the vertical base width \(d(x)=4-x^2\). 2. A square with diagonal \(d\) has area \(A=\frac{d^2}{2}\). 3. Therefore, \(V=\frac{1}{2}\int_{-2}^{2}(4-x^2)^2\,\mathrm{d}x=\frac{256}{15}\).

\(\frac{256}{15}\) cubic units

- What is the horizontal width of the base at height \(y\)? - Convert circumference to radius before calculating circle area. - Integrate with respect to \(y\) over the full height of the base.

1. At height \(y\), the horizontal width is \(w(y)=4-y\). 2. If a circle's circumference is \(w(y)\), then its radius is \(r(y)=\frac{4-y}{2\pi}\). 3. The cross-sectional area is \(A(y)=\pi[r(y)]^2=\frac{(4-y)^2}{4\pi}\). 4. Therefore, \(V=\frac{1}{4\pi}\int_0^4(4-y)^2\,\mathrm{d}y=\frac{16}{3\pi}\).

\(\frac{16}{3\pi}\) cubic units

- Read the fixed diagonal length from the representative cross section. - How can the diagonal and the two rectangle side lengths be related? - Build the rectangular area function before integrating.

1. One side of a cross-sectional rectangle is \(x\). The diagram gives diagonal length \(5\). If the other side is \(h(x)\), then \(x^2+[h(x)]^2=25\), so \(h(x)=\sqrt{25-x^2}\). 2. The cross-sectional area is \(A(x)=x\sqrt{25-x^2}\). 3. The volume is \(V=\int_0^3x\sqrt{25-x^2}\,\mathrm{d}x\). 4. Evaluating gives \(V=\frac{1}{3}(125-64)=\frac{61}{3}\).

\(\frac{61}{3}\) cubic units

- Rewrite both base curves as x-functions of y. - Use the horizontal base width as the semicircle diameter. - Express semicircle area in terms of diameter before integrating.

1. Rewrite the boundaries as \(x=y\) and \(x=\sqrt y\). 2. The horizontal diameter is \(d(y)=\sqrt y-y\) for \(0\le y\le1\). 3. Therefore, \(V=\frac{\pi}{8}\int_0^1(\sqrt y-y)^2\,\mathrm{d}y=\frac{\pi}{240}\).

\(\frac{\pi}{240}\) cubic units

- Solve the circle equation for the upper and lower y-values. - Use the full vertical chord, not one radius, as the square side. - Square the chord length before integrating across the disk.

1. At position \(x\), the disk's vertical chord has length \(s(x)=2\sqrt{9-x^2}\). 2. The square area is \(A(x)=4(9-x^2)\). 3. Therefore, \(V=\int_{-3}^{3}4(9-x^2)\,\mathrm{d}x=144\).

\(144\) cubic units

- Solve the circle equation for both vertical endpoints of a chord. - Use the full chord as the semicircle diameter. - Simplify the squared radical before integrating.

1. The chord length is \(d(x)=2\sqrt{4-x^2}\). 2. The semicircle area is \(A(x)=\frac{\pi}{8}d(x)^2=\frac{\pi}{2}(4-x^2)\). 3. Therefore, \(V=\frac{\pi}{2}\int_{-2}^{2}(4-x^2)\,\mathrm{d}x=\frac{16\pi}{3}\).

\(\frac{16\pi}{3}\) cubic units

- Determine the vertical width of the triangular base at a general x-value. - That width is a square side length, so how is \(A(x)\) related to it? - After integrating, compare the dimensions with a familiar pyramid formula.

1. The slanted boundary through \((0,3)\) and \((3,0)\) is \(y=3-x\), so each square has side length \(s(x)=3-x\) for \(0\le x\le3\). 2. Therefore, \(A(x)=(3-x)^2\). 3. Using cross-sectional integration, \(V=\int_0^3(3-x)^2\,\mathrm{d}x=9\). 4. The linearly shrinking square sections form a square pyramid with base side \(3\) and height \(3\). Its geometric formula gives \(\frac{1}{3}(3^2)(3)=9\), confirming the integral.

a) \(A(x)=(3-x)^2\) b) \(V=\int_0^3(3-x)^2\,\mathrm{d}x=9\) cubic units c) The solid is a square pyramid, and \(\frac{1}{3}(3^2)(3)=9\) cubic units.

- What is the area formula for an equilateral triangle in terms of its side length? - Substitute the given width function before forming the volume integral. - How can \(\sin^2x\) be integrated over \([0,\pi]\)?

1. The area of an equilateral triangle with side \(w\) is \(\frac{\sqrt3}{4}w^2\), so \(A(x)=\frac{\sqrt3}{4}(2+\sin x)^2\). 2. The volume is \(V=\frac{\sqrt3}{4}\int_0^\pi(2+\sin x)^2\,\mathrm{d}x\). 3. Expand and integrate: \(\int_0^\pi(4+4\sin x+\sin^2x)\,\mathrm{d}x=4\pi+8+\frac{\pi}{2}=8+\frac{9\pi}{2}\). 4. Therefore, \(V=2\sqrt3+\frac{9\pi\sqrt3}{8}\,\text{ft}^3\).

a) \(A(x)=\frac{\sqrt3}{4}(2+\sin x)^2\) b) \(2\sqrt3+\frac{9\pi\sqrt3}{8}\,\text{ft}^3\)

- What area corresponds to each measured square side length? - Use the actual interval widths \(1\), \(2\), and \(2\). - Apply the trapezoidal rule to cross-sectional area, not directly to side length.

1. Square the side data to obtain cross-sectional areas \(0\), \(4\), \(9\), and \(1\). 2. Apply trapezoids to \(A(x)\), not to \(s(x)\): \(\frac{1}{2}(0+4)+\frac{2}{2}(4+9)+\frac{2}{2}(9+1)=2+13+10\). 3. Therefore, \(V\approx25\) cubic units.

\(V\approx25\) cubic units

- Convert the side lengths to square areas before using the trapezoidal rule. - Each subinterval has width \(2\). - Use the positivity condition after solving for \(m^2\).

1. The cross-sectional area values are \(1\), \(m^2\), and \(9\). 2. With subinterval width \(2\), the trapezoidal estimate is \((1+m^2)+(m^2+9)=10+2m^2\). 3. Set \(10+2m^2=20\). Then \(m^2=5\), so \(m=\sqrt5\) because \(m>0\).

\(m=\sqrt5\)

- Which trigonometric curve is above the other on the stated interval? - The square side is the vertical distance between the curves. - Simplify the squared difference before integrating.

1. On the interval, \(\cos x\ge\sin x\), so the square side length is \(s(x)=\cos x-\sin x\). 2. The cross-sectional area is \(A(x)=(\cos x-\sin x)^2=1-\sin(2x)\). 3. Therefore, \(V=\int_0^{\pi/4}[1-\sin(2x)]\,\mathrm{d}x=\frac{\pi}{4}-\frac{1}{2}\).

\(\frac{\pi}{4}-\frac{1}{2}\) cubic units

- Determine which base curve is above on \([0,1]\). - Use their vertical separation as the square side length. - Set up the exact integral before using numerical technology.

1. Since \(e^x\ge1+x\), the square side length is \(s(x)=e^x-1-x\). 2. The cross-sectional area is \(A(x)=(e^x-1-x)^2\), so \(V=\int_0^1(e^x-1-x)^2\,\mathrm{d}x\). 3. Numerical evaluation gives \(V\approx0.0913\) cubic units.

\(\approx0.0913\) cubic units

- Read the square diagonal as the vertical separation in the base. - Express a square's area in terms of its diagonal. - Use symmetry to remove the absolute value before integrating.

1. The upper boundary is \(y=2\) and the lower boundary is \(y=|x|\), so the square diagonal is \(d(x)=2-|x|\) on \([-2,2]\). 2. A square with diagonal \(d\) has area \(\frac{1}{2}d^2\). Thus, \(A(x)=\frac{1}{2}(2-|x|)^2\). 3. By symmetry, \(V=2\int_0^2\frac{1}{2}(2-x)^2\,\mathrm{d}x=\frac{8}{3}\) cubic units.

\(\frac{8}{3}\) cubic units

- Use the perimeter condition to recover the missing rectangle dimension. - Verify both dimensions stay nonnegative on the interval.

1. If the other side is \(h(x)\), then \(2[x+h(x)]=12\), so \(h(x)=6-x\). 2. The area is \(A(x)=x(6-x)\). 3. \(V=\int_0^4x(6-x)\,\mathrm{d}x=\frac{80}{3}\).

\(\frac{80}{3}\) cubic units

- What is the vertical distance from the x-axis to the logarithmic curve? - Square that distance to form each square cross-sectional area. - Evaluate the logarithmic antiderivative at \(x=1\) and \(x=e\).

1. The square side length is the vertical distance from the x-axis to the curve: \(s(x)=\ln x\). 2. The cross-sectional area is \(A(x)=(\ln x)^2\), so \(V=\int_1^e(\ln x)^2\,\mathrm{d}x\). 3. An antiderivative is \(x[(\ln x)^2-2\ln x+2]\). Therefore, \(V=e-2\) cubic units.

\(e-2\) cubic units

- Build a linear function from the two endpoint side lengths. - Convert side length into equilateral-triangle area before integrating. - Check that your side function gives the stated endpoint values.

1. The side length has slope \(\frac{2-6}{10}=-\frac{2}{5}\), so \(s(x)=6-\frac{2}{5}x\). 2. The cross-sectional area is \(A(x)=\frac{\sqrt3}{4}\left(6-\frac{2}{5}x\right)^2\). 3. The volume is \(V=\frac{\sqrt3}{4}\int_0^{10}\left(6-\frac{2}{5}x\right)^2\,\mathrm{d}x\). 4. The integral equals \(\frac{520}{3}\), so \(V=\frac{130\sqrt3}{3}\,\text{m}^3\).

a) \(s(x)=6-\frac{2}{5}x\) b) \(\frac{130\sqrt3}{3}\,\text{m}^3\)

- The aspect ratio changes with x, so keep its factor inside the integral. - Both rectangle dimensions include the base width.

1. The cross-sectional area is \(A(x)=w(x)[(1+x)w(x)]=(1+x)(1-x)^2\). 2. \(V=\int_0^1(1+x)(1-x)^2\,\mathrm{d}x=\frac{5}{12}\).

\(\frac{5}{12}\) cubic units

- Differentiate accumulated volume to obtain cross-sectional area. - Take the positive square root of the square's area to obtain its side length. - Minimize the quadratic area expression before taking its square root.

1. The cross-sectional area is \(V'(x)=3x^2-6x+4\). 2. Since the sections are squares, \(s(x)=\sqrt{3x^2-6x+4}\). 3. Rewrite the area as \(3(x-1)^2+1\), which is minimized at \(x=1\). 4. The shortest side has length \(s(1)=1\).

\(s(x)=\sqrt{3x^2-6x+4}\); the shortest side is \(1\) unit at \(x=1\).

- Use the horizontal base width as the square side. - Calculate half of the total volume before forming the cut-height equation. - Solve the accumulation equation numerically within \(0<c<2\).

1. At height \(y\), the square side is \(4-y^2\), so \(A(y)=(4-y^2)^2\). 2. The total volume is \(\int_0^2(4-y^2)^2\,\mathrm{d}y=\frac{256}{15}\). 3. Equal volumes require \(\int_0^c(4-y^2)^2\,\mathrm{d}y=\frac{128}{15}\). 4. Numerical solution on \((0,2)\) gives \(c\approx0.562\).

\(c\approx0.562\)

- Identify where the lower boundary changes from the slanted line to the x-axis. - On each interval, write vertical width as upper boundary minus lower boundary. - Square each width before integrating, then add the two volume contributions.

1. For \(-1\le x\le0\), the upper boundary is \(y=4-x^2\) and the lower boundary is \(y=-3x\), so the square side is \(s_1(x)=4+3x-x^2\). 2. For \(0\le x\le2\), the lower boundary is the x-axis, so the side is \(s_2(x)=4-x^2\). 3. Therefore, \(V=\int_{-1}^{0}(4+3x-x^2)^2\,\mathrm{d}x+\int_{0}^{2}(4-x^2)^2\,\mathrm{d}x\). 4. The two integrals are \(\frac{181}{30}\) and \(\frac{256}{15}\), respectively, so \(V=\frac{231}{10}\) cubic units.

\(s_1(x)=4+3x-x^2\) for \(-1\le x\le0\) \(s_2(x)=4-x^2\) for \(0\le x\le2\) \(V=\frac{231}{10}\) cubic units

- Compare a cross-sectional square with the base square using similarity from the apex. - How does side length scale with distance along the pyramid's height? - Integrate cross-sectional area from the apex to the base, then compare with the standard pyramid formula.

1. The full height is \(9\,\text{cm}\) and the base side length is \(6\,\text{cm}\). By similarity, a square at distance \(z\) from the apex has side length \(s(z)=\frac{6}{9}z=\frac{2}{3}z\). 2. Thus, \(A(z)=s(z)^2=\frac{4}{9}z^2\). 3. The cross-sectional integral gives \(V=\int_0^9\frac{4}{9}z^2\,\mathrm{d}z=108\,\text{cm}^3\). 4. The base area is \(B=6^2=36\,\text{cm}^2\), and \(\frac{1}{3}Bh=\frac{1}{3}(36)(9)=108\,\text{cm}^3\), so the formulas agree.

\(A(z)=\frac{4}{9}z^2\), and \(V=108\,\text{cm}^3\).