Free Math Worksheets

- What is the cross-sectional shape perpendicular to the x-axis? - Identify the outer and inner radii. - Expand the squared binomial carefully.

1. The cross-sections are washers with outer radius \(f(x)\) and inner radius \(g(x)\), so \(V=\pi\int_0^4((\sqrt{x}+1)^2-1^2)\,\text{d}x\). 2. Simplifying gives \(V=\pi\int_0^4(x+2\sqrt{x})\,\text{d}x=\pi[\frac{1}{2}x^2+\frac{4}{3}x^{3/2}]_0^4=\frac{56\pi}{3}\). 3. The comparison integral is \(V_{\text{incorrect}}=\pi\int_0^4(\sqrt{x})^2\,\text{d}x=\pi\int_0^4x\,\text{d}x=8\pi\). 4. The values differ because the area of a washer is \(\pi(R^2-r^2)\), not \(\pi(R-r)^2\).

a) \(V=\frac{56\pi}{3}\) cubic units b) \(V_{\text{incorrect}}=8\pi\) cubic units. It is smaller because \(R^2-r^2\ne(R-r)^2\).

- Determine which function gives the outer radius on the interval. - Use the washer-method formula. - Simplify the integrand before integrating. - Expand the squared binomial carefully.

1. On \([1,3]\), \(f(x)\ge g(x)\ge0\), so the washer method gives \(V=\pi\int_1^3(f(x)^2-g(x)^2)\,\text{d}x\). 2. \(V=\pi\int_1^3((x+1)^2-(\sqrt{2x+2})^2)\,\text{d}x=\pi\int_1^3(x^2-1)\,\text{d}x\). 3. \(V=\pi[\frac{x^3}{3}-x]_1^3=\pi\left(6+\frac{2}{3}\right)=\frac{20\pi}{3}\).

\(V=\frac{20\pi}{3}\) cubic units

- Use the difference of the squares of the outer and inner radii. - Rewrite reciprocal powers with negative exponents before integrating. - Multiply the integral by \(\pi\).

1. On \([1,2]\), \(f(x)\ge g(x)\ge0\), so \(f\) is the outer radius and \(g\) is the inner radius. 2. \(V=\pi\int_1^2\left(\frac{1}{x^2}-\frac{1}{x^4}\right)\,\text{d}x=\pi\int_1^2(x^{-2}-x^{-4})\,\text{d}x\). 3. \(V=\pi[-x^{-1}+\frac{1}{3}x^{-3}]_1^2=\pi\left(-\frac{11}{24}+\frac{2}{3}\right)=\frac{5\pi}{24}\).

\(V=\frac{5\pi}{24}\) cubic units

- Use the disc-method formula for the exterior volume. - Square the interior-radius function before integrating. - Subtract the interior volume from the exterior volume. - The radius at a given \(x\) is the function value there.

1. The exterior volume is \(V_f=\pi\int_0^{10}(\sqrt{x+2})^2\,\text{d}x=\pi\int_0^{10}(x+2)\,\text{d}x=70\pi\,\text{cm}^3\). 2. The interior volume is \(V_g=\pi\int_0^{10}(0.2x+1)^2\,\text{d}x=\frac{130\pi}{3}\,\text{cm}^3\). 3. The glass volume is \(V_f-V_g=70\pi-\frac{130\pi}{3}=\frac{80\pi}{3}\,\text{cm}^3\). 4. Solve \(\sqrt{x+2}=3\). Squaring gives \(x+2=9\), so \(x=7\,\text{cm}\).

a) \(70\pi\,\text{cm}^3\) b) \(\frac{130\pi}{3}\,\text{cm}^3\) c) \(\frac{80\pi}{3}\,\text{cm}^3\) d) \(x=7\,\text{cm}\)

- First find the container volume up to a variable water depth. - A solid object reduces the space available for water. - Determine whether the cylinder is partially or fully submerged at each depth. - Subtract the displaced cylinder volume from the container volume.

1. The container volume up to height \(h\) is \(V(h)=\pi\int_0^h(f(x))^2\,\text{d}x=\pi\int_0^h(2x+4)\,\text{d}x=\pi(h^2+4h)\). 2. At \(h=2\), the container volume is \(12\pi\,\text{in.}^3\). The cylinder occupies \(\pi(1)^2(2)=2\pi\,\text{in.}^3\), so the water volume is \(10\pi\,\text{in.}^3\approx31.42\,\text{in.}^3\). 3. At \(h=5\), the container volume is \(45\pi\,\text{in.}^3\). The entire cylinder remains submerged and displaces \(2\pi\,\text{in.}^3\), so the water volume is \(43\pi\,\text{in.}^3\approx135.09\,\text{in.}^3\).

a) \(10\pi\,\text{in.}^3\approx31.42\,\text{in.}^3\) b) \(43\pi\,\text{in.}^3\approx135.09\,\text{in.}^3\)

- Find the intersection points first. - Determine which function gives the outer radius. - Use the washer-method formula. - Expand the squared expressions carefully.

1. The curves intersect where \(x^2=4x-x^2\), giving \(x=0\) and \(x=2\). 2. On \([0,2]\), \(g(x)\ge f(x)\ge0\), so the washer method gives \(V=\pi\int_0^2(g(x)^2-f(x)^2)\,\text{d}x\). 3. \((4x-x^2)^2-(x^2)^2=16x^2-8x^3\). 4. \(V=\pi[\frac{16}{3}x^3-2x^4]_0^2=\frac{32\pi}{3}\).

\(V=\frac{32\pi}{3}\) cubic units

- Find the intersection points to determine the bounds. - Use the difference of the squares of the outer and inner radii. - Rewrite reciprocal powers using negative exponents. - Check that the resulting volume is positive.

1. The curves intersect where \(\frac{2}{x}=3-x\), which gives \(x^2-3x+2=0\). Thus, the bounds are \(x=1\) and \(x=2\). 2. On \([1,2]\), \(g(x)\ge f(x)>0\), so \(V=\pi\int_1^2\left((3-x)^2-\left(\frac{2}{x}\right)^2\right)\,\text{d}x\). 3. \(V=\pi\int_1^2(x^2-6x+9-4x^{-2})\,\text{d}x\). 4. \(V=\pi[\frac{x^3}{3}-3x^2+9x+\frac{4}{x}]_1^2=\pi\left(\frac{32}{3}-\frac{31}{3}\right)=\frac{\pi}{3}\).

\(V=\frac{\pi}{3}\) cubic units

- View the bowl as an outer solid with an inner solid removed. - Use the disc-method formula for each surface of revolution. - Pay attention to the different intervals for the outer and inner surfaces. - The capacity is the volume of the interior cavity.

1. The outer volume is \(V_{\text{outer}}=\pi\int_0^{16}(0.5x+4)\,\text{d}x\). 2. \(V_{\text{outer}}=\pi[0.25x^2+4x]_0^{16}=128\pi\,\text{cm}^3\). 3. The capacity is the inner volume: \(V_{\text{inner}}=\pi\int_4^{16}(0.8x-3.2)\,\text{d}x\). 4. \(V_{\text{inner}}=\pi[0.4x^2-3.2x]_4^{16}=57.6\pi\approx180.96\,\text{cm}^3\). 5. The glass volume is \(V_{\text{outer}}-V_{\text{inner}}=(128-57.6)\pi=70.4\pi\approx221.17\,\text{cm}^3\).

The glass volume is \(70.4\pi\,\text{cm}^3\approx221.17\,\text{cm}^3\). The capacity is \(57.6\pi\,\text{cm}^3\approx180.96\,\text{cm}^3\).

- Subtract the inner cross-sectional area from the outer cross-sectional area. - Apply exponent rules when squaring each radius function. - Use the antiderivative of \(e^{kx}\). - The two integrals can be combined because they use the same interval.

1. Use washers: \(V=\pi\int_0^{10}(f(x)^2-g(x)^2)\,\text{d}x\). 2. \(f(x)^2-g(x)^2=4e^{0.2x}-2.25e^{0.2x}=1.75e^{0.2x}\). 3. Therefore, \(V=1.75\pi\int_0^{10}e^{0.2x}\,\text{d}x=1.75\pi[5e^{0.2x}]_0^{10}\). 4. Thus, \(V=8.75\pi(e^2-1)\approx175.63\,\text{cm}^3\).

\(8.75\pi(e^2-1)\,\text{cm}^3\approx175.63\,\text{cm}^3\)

- Find the intersection points to determine the limits of integration. - Identify which curve gives the outer radius. - Use the washer-method formula. - Subtract the squares of the radii before integrating.

1. The curves intersect where \(2\sqrt{x}=x\). Squaring gives \(4x=x^2\), so \(x=0\) or \(x=4\). 2. On \([0,4]\), \(f(x)\ge g(x)\). Using washers, \(V=\pi\int_0^4((2\sqrt{x})^2-x^2)\,\text{d}x\). 3. \(V=\pi\int_0^4(4x-x^2)\,\text{d}x=\pi\left[2x^2-\frac{x^3}{3}\right]_0^4\). 4. Therefore, \(V=\pi\left(32-\frac{64}{3}\right)=\frac{32\pi}{3}\approx33.51\) cubic units.

\(\frac{32\pi}{3}\) cubic units, or approximately \(33.51\) cubic units

- Find the intersection points in the first quadrant. - Use a power-reduction identity for \(\sin^2x\). - Apply the washer-method formula and multiply the entire difference by \(\pi\).

1. The curves intersect at \(x=0\) and \(x=\frac{\pi}{2}\). 2. On \([0,\frac{\pi}{2}]\), \(\sin x\) is the outer radius. Thus, \(V=\pi\int_0^{\pi/2}\left(\sin^2x-\left(\frac{2x}{\pi}\right)^2\right)\,\text{d}x\). 3. Using \(\int\sin^2x\,\text{d}x=\frac{x}{2}-\frac{\sin(2x)}{4}\), \(\int_0^{\pi/2}\sin^2x\,\text{d}x=\frac{\pi}{4}\). 4. Also, \(\int_0^{\pi/2}\frac{4x^2}{\pi^2}\,\text{d}x=\frac{\pi}{6}\). 5. Therefore, \(V=\pi\left(\frac{\pi}{4}-\frac{\pi}{6}\right)=\frac{\pi^2}{12}\approx0.822\) cubic units.

\(\frac{\pi^2}{12}\) cubic units, or approximately \(0.822\) cubic units

- Solve for the intersection points and check any solutions introduced by squaring. - Identify which function gives the outer radius. - Use the washer-method formula. - Subtract the squares of the functions, not the square of their difference.

1. Set the functions equal: \(\sqrt{x+4}=\frac{1}{2}x+2\). Squaring and checking the solutions gives intersections at \(x=-4\) and \(x=0\). 2. On \([-4,0]\), \(f(x)\ge g(x)\), and both functions are nonnegative. Thus, \(V=\pi\int_{-4}^{0}\left((\sqrt{x+4})^2-\left(\frac{x}{2}+2\right)^2\right)\,\text{d}x\). 3. The integrand simplifies to \(-\frac{x^2}{4}-x\). 4. \(V=\pi\left[-\frac{x^3}{12}-\frac{x^2}{2}\right]_{-4}^{0}=\pi\left(\frac{8}{3}\right)=\frac{8\pi}{3}\approx8.38\) cubic units.

\(\frac{8\pi}{3}\) cubic units, or approximately \(8.38\) cubic units

- Find the intersection points to determine the interval. - Identify the outer and inner radii. - Square each function separately before subtracting. - A sketch can help confirm which graph is farther from the x-axis.

1. Solve \(x^2-4x+5=-x+5\). This gives \(x(x-3)=0\), so the curves intersect at \(x=0\) and \(x=3\). 2. On \([0,3]\), \(g(x)\ge f(x)>0\), so \(g\) is the outer radius and \(f\) is the inner radius. 3. \(V=\pi\int_0^3(g(x)^2-f(x)^2)\,\text{d}x\). 4. The difference is \((-x+5)^2-(x^2-4x+5)^2=-x^4+8x^3-25x^2+30x\). 5. \(V=\pi\left[-\frac{x^5}{5}+2x^4-\frac{25x^3}{3}+15x^2\right]_0^3=\frac{117\pi}{5}\approx73.51\) cubic units.

\(\frac{117\pi}{5}\) cubic units, or approximately \(73.51\) cubic units

- Find the intersection points first. - For area, integrate the upper function minus the lower function. - For volume, identify the outer and inner radii. - Square each radius function before subtracting.

1. Solving \(\frac{1}{2}x^2=x+4\) gives intersection points \(x=-2\) and \(x=4\). 2. On \([-2,4]\), \(g(x)\ge f(x)\). The area is \(A=\int_{-2}^{4}\left(x+4-\frac{x^2}{2}\right)\,\text{d}x=18\) square units. 3. Both functions are nonnegative on the interval, so washers have outer radius \(g(x)\) and inner radius \(f(x)\). 4. \(V=\pi\int_{-2}^{4}\left((x+4)^2-\left(\frac{x^2}{2}\right)^2\right)\,\text{d}x\). 5. \(V=\pi\left[\frac{x^3}{3}+4x^2+16x-\frac{x^5}{20}\right]_{-2}^{4}=\frac{576\pi}{5}\approx361.91\) cubic units.

a) \(18\) square units b) \(\frac{576\pi}{5}\) cubic units, or approximately \(361.91\) cubic units

- Set the functions equal to find the limits. - Use the logarithmic antiderivative for \(1/x\). - For the volume, subtract the squares of the radii. - Rewrite \(1/x^2\) as a negative power before integrating.

1. Solving \(4-x=\frac{3}{x}\) gives \(x^2-4x+3=0\), so the curves intersect at \(x=1\) and \(x=3\). 2. On \([1,3]\), \(h(x)\ge k(x)\). Thus, \(A=\int_1^3\left(4-x-\frac{3}{x}\right)\,\text{d}x\). 3. \(A=\left[4x-\frac{x^2}{2}-3\ln x\right]_1^3=4-3\ln3\approx0.704\) square units. 4. Using washers, \(V=\pi\int_1^3\left((4-x)^2-\left(\frac{3}{x}\right)^2\right)\,\text{d}x\). 5. \(V=\pi\left[16x-4x^2+\frac{x^3}{3}+\frac{9}{x}\right]_1^3=\frac{8\pi}{3}\approx8.378\) cubic units.

a) \(4-3\ln3\) square units, or approximately \(0.704\) square units b) \(\frac{8\pi}{3}\) cubic units, or approximately \(8.378\) cubic units

- Find the intersection points first. - Identify the outer and inner radii. - Expand both squared binomials carefully. - Symmetry can shorten the calculation.

1. Solving \(x^2+1=9-x^2\) gives \(x=-2\) and \(x=2\). 2. On \([-2,2]\), \(g(x)\ge f(x)>0\), so washers have outer radius \(g(x)\) and inner radius \(f(x)\). 3. \(V=\pi\int_{-2}^{2}\left((9-x^2)^2-(x^2+1)^2\right)\,\text{d}x\). 4. The integrand simplifies to \(80-20x^2\). 5. Thus, \(V=\pi\left[80x-\frac{20x^3}{3}\right]_{-2}^{2}=\frac{640\pi}{3}\approx670.21\) cubic units.

\(\frac{640\pi}{3}\) cubic units, or approximately \(670.21\) cubic units

- Set the function values equal to find the intersections. - Check solutions introduced by squaring. - Identify the outer radius on the interval. - Use the washer-method formula.

1. Set the functions equal: \(\sqrt{2x}=x\). Squaring and checking gives \(x=0\) and \(x=2\), so the intersection points are \((0,0)\) and \((2,2)\). 2. On \([0,2]\), \(f(x)\ge g(x)\). Using washers, \(V=\pi\int_0^2(2x-x^2)\,\text{d}x\). 3. \(V=\pi\left[x^2-\frac{x^3}{3}\right]_0^2=\frac{4\pi}{3}\).

a) \((0,0)\) and \((2,2)\) b) \(\frac{4\pi}{3}\) cubic units

- Use the disc method for the cavity volume. - Subtract the cavity volume from the full outer volume. - Pay attention to the different intervals. - Use mass equals density times volume.

1. The cavity volume is \(V_{\text{cavity}}=\pi\int_1^6(4x-4)\,\text{d}x=50\pi\approx157.1\,\text{cm}^3=157.1\,\text{mL}\). 2. The full outer volume is \(V_{\text{outer}}=\pi\int_0^6(0.5x+3)^2\,\text{d}x=126\pi\,\text{cm}^3\). 3. The glass volume is \(126\pi-50\pi=76\pi\approx238.8\,\text{cm}^3\). 4. The mass is \(2.5(76\pi)=190\pi\approx596.9\,\text{g}\).

a) \(157.1\,\text{mL}\) b) \(76\pi\,\text{cm}^3\approx238.8\,\text{cm}^3\) c) \(596.9\,\text{g}\)

- Evaluate the radius function at the endpoints and the midpoint. - Use the disc-method formula for the solid spool. - Subtract the cylindrical hole from the solid volume. - Convert the given hole diameter to a radius.

1. At both ends, \(f(0)=f(8)=\sqrt{12}\), so each end diameter is \(2\sqrt{12}=4\sqrt3\approx6.93\,\text{cm}\). At \(x=4\), the radius is \(2\), so the middle diameter is \(4\,\text{cm}\). 2. By the disc method, \(V_{\text{solid}}=\pi\int_0^8(0.5(x-4)^2+4)\,\text{d}x=\frac{160\pi}{3}\approx167.55\,\text{cm}^3\). 3. The hole has radius \(1\,\text{cm}\) and length \(8\,\text{cm}\), so its volume is \(8\pi\,\text{cm}^3\). 4. The wood volume is \(\frac{160\pi}{3}-8\pi=\frac{136\pi}{3}\approx142.42\,\text{cm}^3\).

a) End diameters: \(4\sqrt3\,\text{cm}\approx6.93\,\text{cm}\); middle diameter: \(4\,\text{cm}\) b) \(\frac{160\pi}{3}\,\text{cm}^3\approx167.55\,\text{cm}^3\) c) \(\frac{136\pi}{3}\,\text{cm}^3\approx142.42\,\text{cm}^3\)

- Identify the outer and inner radii. - Use the washer-method formula. - Square each radius before subtracting. - Simplify the squared radical.

1. On \([0,16]\), the outer radius is \(4\) and the inner radius is \(\sqrt{x}\). 2. Using washers, \(V=\pi\int_0^{16}(4^2-(\sqrt{x})^2)\,\text{d}x=\pi\int_0^{16}(16-x)\,\text{d}x\). 3. \(V=\pi\left[16x-\frac{x^2}{2}\right]_0^{16}=128\pi\approx402.12\) cubic units.

\(128\pi\) cubic units, or approximately \(402.12\) cubic units

- Decide which function gives the outer radius. - Use the washer-method formula. - Rewrite \(1/x^2\) with a negative exponent before integrating.

1. On \([1,2]\), \(f(x)\ge g(x)\), so washers have outer radius \(1/x\) and inner radius \(0.5\). 2. \(V=\pi\int_1^2\left(\frac{1}{x^2}-0.25\right)\,\text{d}x\). 3. \(V=\pi\left[-\frac{1}{x}-0.25x\right]_1^2=\frac{\pi}{4}\approx0.785\) cubic units.

\(\frac{\pi}{4}\) cubic units, or approximately \(0.785\) cubic units

- Find the intersections in the first quadrant. - Identify the outer radius on \([0,1]\). - Use exponent rules when squaring the functions.

1. Solving \(x^2=x^4\) in the first quadrant gives \(x=0\) and \(x=1\). 2. On \([0,1]\), \(x^2\ge x^4\). Using washers, \(V=\pi\int_0^1(x^4-x^8)\,\text{d}x\). 3. \(V=\pi\left[\frac{x^5}{5}-\frac{x^9}{9}\right]_0^1=\pi\left(\frac{1}{5}-\frac{1}{9}\right)=\frac{4\pi}{45}\approx0.279\) cubic units.

\(\frac{4\pi}{45}\) cubic units, or approximately \(0.279\) cubic units

- Find the intersection points. - Identify the outer and inner radii. - Square both the coefficient and variable in \(2x\). - Use the washer-method formula.

1. Solving \(2x=x^2\) gives \(x=0\) and \(x=2\). 2. On \([0,2]\), \(2x\ge x^2\). Using washers, \(V=\pi\int_0^2(4x^2-x^4)\,\text{d}x\). 3. \(V=\pi\left[\frac{4x^3}{3}-\frac{x^5}{5}\right]_0^2=\frac{64\pi}{15}\approx13.40\) cubic units.

\(\frac{64\pi}{15}\) cubic units, or approximately \(13.40\) cubic units

- Find the intersections first. - Identify the outer and inner radii. - Expand the squared binomial carefully. - Use symmetry to simplify the integral.

1. Solving \(-x^2+5=1\) gives \(x=-2\) and \(x=2\). 2. On \([-2,2]\), \(f(x)\ge g(x)\), so washers have outer radius \(f(x)\) and inner radius \(1\). 3. \(V=\pi\int_{-2}^{2}(({-x^2+5})^2-1)\,\text{d}x\). 4. The integrand is \(x^4-10x^2+24\). Using symmetry, \(V=2\pi\left[\frac{x^5}{5}-\frac{10x^3}{3}+24x\right]_0^2=\frac{832\pi}{15}\approx174.25\) cubic units.

\(\frac{832\pi}{15}\) cubic units, or approximately \(174.25\) cubic units

- Identify the outer radius. - Use the washer-method formula. - Squaring the square root simplifies the integrand. - Square the linear inner radius before subtracting.

1. The curves meet at \(x=0\) and \(x=4\), and \(\sqrt{x}\ge0.5x\) on this interval. 2. Using washers, \(V=\pi\int_0^4\left(x-0.25x^2\right)\,\text{d}x\). 3. \(V=\pi\left[\frac{x^2}{2}-\frac{x^3}{12}\right]_0^4=\frac{8\pi}{3}\approx8.38\) cubic units.

\(\frac{8\pi}{3}\) cubic units, or approximately \(8.38\) cubic units

- Identify which graph bounds the hollow interior and which graph gives the outside radius. - Use the volume formula for a solid formed by rotation about the x-axis. - Account for the different intervals on which the inner and outer radii are defined. - Interpret each function value as a radius at a particular position.

1. The capacity is the volume of the inner solid of revolution: \(V_{\text{inside}}=\pi\int_1^4(2\sqrt{x-1})^2\,dx=\pi\int_1^4 4(x-1)\,dx=18\pi\approx56.55\,\text{cm}^3\). 2. The total outer volume is \(V_{\text{outer}}=\pi\int_0^4(0.2x+3)^2\,dx=\frac{3484}{75}\pi\approx145.94\,\text{cm}^3\). 3. The volume of glass is \(V_{\text{glass}}=V_{\text{outer}}-V_{\text{inside}}=\frac{2134}{75}\pi\approx89.39\,\text{cm}^3\). 4. Using the unrounded volume, the mass is \(m=2.5\cdot\frac{2134}{75}\pi=\frac{1067}{15}\pi\approx223.47\,\text{g}\). 5. At the top, \(f(4)-g(4)=3.8-2\sqrt{3}\approx0.34\,\text{cm}\). This is the wall thickness at the rim.

a) \(18\pi\approx56.55\,\text{cm}^3\) b) \(\frac{2134}{75}\pi\approx89.39\,\text{cm}^3\) c) \(\frac{1067}{15}\pi\approx223.47\,\text{g}\) d) It is the wall thickness at the top rim, about \(0.34\,\text{cm}\).

- Write the area of a washer as the area of the outer circle minus the inner circle. - Compare that expression with the expansion of \((a-b)^2\). - Under what conditions are \(a^2-b^2\) and \((a-b)^2\) equal? - What does an inner radius of zero mean geometrically?

1. The actual washer area is \(A(x)=\pi(f(x)^2-g(x)^2)\). 2. The proposed expression is \(\pi(f(x)-g(x))^2=\pi(f(x)^2-2f(x)g(x)+g(x)^2)\). 3. Setting the expressions equal and simplifying gives \(2g(x)(f(x)-g(x))=0\). 4. Therefore, equality holds only when \(g(x)=0\), so the cross-section is a disc, or when \(f(x)=g(x)\), so the cross-sectional area is zero. 5. For a genuine washer with \(f(x)>g(x)>0\), the correct integrand is \(\pi(f(x)^2-g(x)^2)\). Squaring the radial thickness \(f(x)-g(x)\) ignores the washer's distance from the axis.

\(\pi(f^2-g^2)-\pi(f-g)^2=2\pi g(f-g)\). This difference is positive when \(f>g>0\), so \(\pi(f-g)^2\) cannot be used for a genuine washer. The correct volume uses \(\pi\int(f^2-g^2)\,\text{d}x\).

- Identify which dimension is represented by the x-axis and which by the function values. - A zero indicates where an interior or exterior boundary begins on the axis. - Gross volume includes everything enclosed by the outer surface. - Use the disc method for the exterior volume. - Subtract the interior volume to find the material volume.

1. The tower's height is the length of the exterior interval, \(60\,\text{m}\). 2. The zeros are \(0\) and \(0.8\), so their difference is \(0.8\,\text{m}\). This is the axial thickness of the base. 3. \(f(60)-g(60)=\sqrt{270}-\sqrt{248.64}\approx0.66\,\text{m}\). This is the wall thickness at the top. 4. The gross volume is \(V_{\text{out}}=\pi\int_0^{60}4.5x\,\text{d}x=8100\pi\approx25{,}446.90\,\text{m}^3\). 5. The interior volume is \(V_{\text{in}}=\pi\int_{0.8}^{60}4.2(x-0.8)\,\text{d}x=7359.744\pi\,\text{m}^3\). Therefore, the material volume is \(V_{\text{out}}-V_{\text{in}}=740.256\pi\approx2325.58\,\text{m}^3\).

a) \(60\,\text{m}\) b) \(0.8\,\text{m}\); the base thickness c) Approximately \(0.66\,\text{m}\); the wall thickness at the top d) \(8100\pi\,\text{m}^3\approx25{,}446.90\,\text{m}^3\) e) \(740.256\pi\,\text{m}^3\approx2325.58\,\text{m}^3\)

- Expand the squared radius before integrating. - Use the point-slope form of a tangent line. - For the open core, subtract the square of the inner radius from the square of the outer radius. - The floor slab is a circle whose radius is the profile value at \(x=0\).

1. By the disc method, \(V=\pi\int_0^5(4-0.1x^2)^2\,\text{d}x\). 2. Expanding and integrating gives \(V=\pi\left[16x-\frac{4}{15}x^3+\frac{1}{500}x^5\right]_0^5=\frac{635\pi}{12}\approx166.24\,\text{m}^3\). 3. \(g(3)=3.1\) and \(g'(3)=-0.6\). The tangent line is \(y-3.1=-0.6(x-3)\), or \(y=-0.6x+4.9\). 4. For \(0\le x\le3\), use washers with outer radius \(g(x)\) and inner radius \(2\): \(V_{\text{between}}=\pi\int_0^3((4-0.1x^2)^2-2^2)\,\text{d}x\). 5. \(V_{\text{between}}=\pi\left[12x-\frac{4}{15}x^3+\frac{1}{500}x^5\right]_0^3=\frac{14{,}643\pi}{500}\approx92.00\,\text{m}^3\). 6. The slab area is \(A=\pi(g(0))^2=16\pi\,\text{m}^2\). The cost is \(45(16\pi)=720\pi\approx\$2261.95\).

a) Approximately \(166.24\,\text{m}^3\) b) \(y=-0.6x+4.9\) c) Approximately \(92.00\,\text{m}^3\) d) Approximately \(\$2261.95\)