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Question 1. Find the volume of revolution formed by rotating the region \(R\), defined by \(y=x^{1/2}\), \(x=4\), and \(y=0\), about the \(y\)-axis.
Use the cylindrical shell method. A shell at \(x\) has radius \(x\), height \(\sqrt{x}\), and thickness \(dx\). Thus \(\displaystyle V=2\pi\int_0^4x\sqrt{x}\,dx=2\pi\int_0^4x^{3/2}\,dx\). Integrating, \(\displaystyle V=2\pi\left[\frac{2}{5}x^{5/2}\right]_0^4=\frac{4\pi}{5}(4^{5/2})\). Since \(4^{5/2}=32\), \(\displaystyle V=\frac{128\pi}{5}\). Answer: C.
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Question 2. Find the volume of the region \(R\) between \(x^{1/2}+y^{1/2}=2\) and the \(x\)-axis from \(x=0\) to \(x=2\), when it is rotated about the \(x\)-axis.
Solve for \(y\): \(\sqrt{y}=2-\sqrt{x}\), so \(y=(2-\sqrt{x})^2\). Using the disk method, \(\displaystyle V=\pi\int_0^2y^2\,dx=\pi\int_0^2(2-\sqrt{x})^4\,dx\). Expand: \((2-\sqrt{x})^4=16-32x^{1/2}+24x-8x^{3/2}+x^2\). Therefore \(\displaystyle V=\pi\left[16x-\frac{64}{3}x^{3/2}+12x^2-\frac{16}{5}x^{5/2}+\frac{x^3}{3}\right]_0^2\). The numerical value is approximately \(13.28\), matching option E. Answer: E.
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Question 3. Which of the following yields the volume of the solid generated by revolving the region bounded by \(y=x^3\) and \(y=x\), from \(x=0\) to \(x=1\), about the \(y\)-axis?
Because the axis of rotation is the \(y\)-axis, use washers with respect to \(y\). Rewrite the curves as \(x=y^{1/3}\) from \(y=x^3\), and \(x=y\) from \(y=x\). For \(0\le y\le1\), the outer radius is \(R(y)=y^{1/3}\), and the inner radius is \(r(y)=y\). Hence \(\displaystyle V=\pi\int_0^1\left(R^2-r^2\right)\,dy=\pi\int_0^1\left(y^{2/3}-y^2\right)\,dy\). Answer: D.
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Question 4. Find the volume generated by rotating about the \(x\)-axis the area bounded by \(x=0\), \(y=2\sin x\cos x\), \(y=2\cos x\), and \(x=\frac{\pi}{2}\).
Use washers on \(0\le x\le\frac{\pi}{2}\). The outer radius is \(R(x)=2\cos x\), and the inner radius is \(r(x)=2\sin x\cos x=\sin2x\). Thus \(\displaystyle V=\pi\int_0^{\pi/2}\left[4\cos^2x-\sin^2(2x)\right]dx\). Using \(\cos^2x=\frac{1+\cos2x}{2}\) and \(\sin^2(2x)=\frac{1-\cos4x}{2}\), the integrand becomes \(\frac32+2\cos2x+\frac12\cos4x\). Therefore \(\displaystyle V=\pi\left[\frac32x+\sin2x+\frac18\sin4x\right]_0^{\pi/2}=\frac{3\pi^2}{4}\). Answer: D.
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Question 5. The base of a solid is the circular region \(x^2+y^2=9\). Every cross section perpendicular to the \(x\)-axis is an equilateral triangle with one side on the base. Find the volume.
For a fixed \(x\), the circle gives \(y=\pm\sqrt{9-x^2}\), so the side length is \(s=2\sqrt{9-x^2}\). The area of an equilateral triangle is \(\displaystyle A=\frac{\sqrt3}{4}s^2\), so \(\displaystyle A(x)=\sqrt3(9-x^2)\). Thus \(\displaystyle V=\int_{-3}^{3}\sqrt3(9-x^2)\,dx=2\sqrt3\int_0^3(9-x^2)\,dx\). Evaluating, \(\displaystyle V=2\sqrt3\left[9x-\frac{x^3}{3}\right]_0^3=36\sqrt3\). Answer: E.
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Question 6. Find the volume of the solid formed by rotating the region enclosed by \(4x^2+9y^2=36\) about the \(x\)-axis.
Solve for \(y^2\): \(\displaystyle 9y^2=36-4x^2\), so \(\displaystyle y^2=4-\frac49x^2\). The ellipse intersects the \(x\)-axis at \(x=\pm3\). Using the disk method, \(\displaystyle V=\pi\int_{-3}^{3}\left(4-\frac49x^2\right)dx\). By symmetry, \(\displaystyle V=2\pi\int_0^3\left(4-\frac49x^2\right)dx=2\pi\left[4x-\frac{4x^3}{27}\right]_0^3=16\pi\). Answer: D.
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Question 7. The region bounded by \(f(x)=x(2-x)\) and the \(x\)-axis is revolved about the \(y\)-axis. Find the volume of the solid.
The curve meets the \(x\)-axis when \(x(2-x)=0\), so \(x=0\) and \(x=2\). Use cylindrical shells about the \(y\)-axis. A shell has radius \(x\) and height \(x(2-x)=2x-x^2\). Thus \(\displaystyle V=2\pi\int_0^2x(2x-x^2)\,dx=2\pi\int_0^2(2x^2-x^3)\,dx\). Therefore \(\displaystyle V=2\pi\left[\frac{2x^3}{3}-\frac{x^4}{4}\right]_0^2=\frac{8\pi}{3}\text{ units}^3\). Answer: C.
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Question 8. The base of a solid is the circular region \(x^2+y^2=9\). Every cross section perpendicular to the \(x\)-axis is a square with one side on the base. Find the volume.
For a fixed \(x\), the side length of the square is the vertical distance across the circle: \(\displaystyle s=2\sqrt{9-x^2}\). Thus the cross-sectional area is \(\displaystyle A(x)=s^2=4(9-x^2)\). The volume is \(\displaystyle V=\int_{-3}^{3}4(9-x^2)\,dx=8\int_0^3(9-x^2)\,dx\). Therefore \(\displaystyle V=8\left[9x-\frac{x^3}{3}\right]_0^3=8(18)=144\). Answer: C.
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Question 9. Let \(R\) be a first-quadrant region enclosed by \(y=(2-x)(2+x)\), \(y=3x\), and the \(y\)-axis. Find the volume generated by revolving \(R\) about the \(x\)-axis.
The curves are \(y=4-x^2\) and \(y=3x\). Find the first-quadrant intersection: \(4-x^2=3x\), so \(x^2+3x-4=0\), giving \(x=1\). Using washers about the \(x\)-axis, \(\displaystyle V=\pi\int_0^1\left[(4-x^2)^2-(3x)^2\right]dx\). Expand the integrand: \((4-x^2)^2-9x^2=x^4-17x^2+16\). Thus \(\displaystyle V=\pi\left[\frac{x^5}{5}-\frac{17x^3}{3}+16x\right]_0^1=\pi\left(\frac15-\frac{17}{3}+16\right)=\frac{158\pi}{15}\). Answer: C.
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Question 10. Find the volume of the solid formed by revolving the region bounded by \(y=\sqrt{x-2}\), \(y=0\), and \(x=6\) about the \(x\)-axis.
The curve meets \(y=0\) when \(\sqrt{x-2}=0\), so the interval is \(2\le x\le6\). Using the disk method, \(\displaystyle V=\pi\int_2^6\left(\sqrt{x-2}\right)^2dx=\pi\int_2^6(x-2)\,dx\). Therefore \(\displaystyle V=\pi\left[\frac{x^2}{2}-2x\right]_2^6=\pi\left(6-(-2)\right)=8\pi\). Answer: A.
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