05. 정적분의 활용Applications of Definite Integrals

Solution. By the disk method, the volume is $$V = \int_0^4 \pi y^2\,dx = \int_0^4 \pi x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi \cdot \frac{16}{2} = 8\pi.$$

Solution. (1) Revolving the disk $x^2+y^2\le a^2$ about the $x$-axis produces a stack of disks of radius $\sqrt{a^2-x^2}$, so $$V = \int_{-a}^{a} \pi (a^2-x^2)\,dx = \pi\left[a^2x - \frac{x^3}{3}\right]_{-a}^{a} = \pi\left(\frac{2a^3}{3}-\left(-\frac{2a^3}{3}\right)\right)=\frac{4}{3}\pi a^3.$$

(2) The curves $y=\sqrt{x}$ and $y=1$ meet at $x=1$, and $y=\sqrt{x}$ lies above the line $y=1$ for $1\le x\le 4$. Revolving about the line $y=1$ produces disks of radius $\sqrt{x}-1$, so $$V = \int_1^4 \pi(\sqrt{x}-1)^2\,dx = \pi\int_1^4 (x-2\sqrt{x}+1)\,dx = \pi\left[\frac{x^2}{2}-\frac{4}{3}x^{3/2}+x\right]_1^4.$$ At $x=4$, $\dfrac{x^2}{2}-\dfrac43x^{3/2}+x = 8-\dfrac{32}{3}+4=\dfrac43$, and at $x=1$, $\dfrac12-\dfrac43+1=\dfrac16$, so $$V = \pi\left(\frac43-\frac16\right)=\frac{7}{6}\pi.$$

Solution. (1) Revolving the curve $x=2/y$, $1\le y\le4$, about the $y$-axis produces disks of radius $2/y$, so $$V = \int_1^4 \pi\left(\frac{2}{y}\right)^2 dy = 4\pi\int_1^4 \frac{1}{y^2}\,dy = 4\pi\left[-\frac1y\right]_1^4 = 4\pi\left(-\frac14+1\right)=3\pi.$$

(2) The curve $x=y^2+1$ and the line $x=3$ meet where $y^2+1=3$, i.e., $y=\pm\sqrt2$. Revolving about the line $x=3$ produces disks of radius $3-(y^2+1)=2-y^2$, so $$V = \int_{-\sqrt2}^{\sqrt2} \pi(2-y^2)^2\,dy = 2\pi\int_0^{\sqrt2}(4-4y^2+y^4)\,dy$$ $$= 2\pi\left[4y-\frac43y^3+\frac{y^5}{5}\right]_0^{\sqrt2} = 2\pi\left(4\sqrt2-\frac{8\sqrt2}{3}+\frac{4\sqrt2}{5}\right)=2\pi\cdot\frac{32\sqrt2}{15}=\frac{64\sqrt2}{15}\pi.$$

Solution. (1) The curve $y=x^2+1$ and the line $y=-x+3$ meet where $x^2+1=-x+3$, i.e., $x^2+x-2=(x+2)(x-1)=0$, so $x=-2,1$. Since the line lies above the curve for $-2\le x\le1$, revolving about the $x$-axis by the washer method gives $$V = \pi\int_{-2}^{1}\left[(3-x)^2-(x^2+1)^2\right]dx = \pi\int_{-2}^1 (8-6x-x^2-x^4)\,dx$$ $$=\pi\left[8x-3x^2-\frac{x^3}{3}-\frac{x^5}{5}\right]_{-2}^1 = \pi\left(\frac{67}{15}-\left(-\frac{284}{15}\right)\right)=\pi\cdot\frac{351}{15}=\frac{117}{5}\pi.$$

(2) The curve $y=x^2$ and the line $y=2x$ meet at $x=0,2$. For $0\le y\le4$, the region satisfies $\dfrac y2\le x\le\sqrt y$, so revolving about the $y$-axis by the washer method gives $$V=\pi\int_0^4\left[(\sqrt y)^2-\left(\frac y2\right)^2\right]dy=\pi\int_0^4\left(y-\frac{y^2}{4}\right)dy=\pi\left[\frac{y^2}{2}-\frac{y^3}{12}\right]_0^4=\pi\left(8-\frac{16}{3}\right)=\frac{8}{3}\pi.$$

(3) By Pappus's theorem, since the centroid of the disk is at distance $b$ from the axis and the disk has area $\pi a^2$, the volume of the torus is $$V = 2\pi b\cdot \pi a^2 = 2\pi^2 a^2 b.$$

Solution. The curve $y=3x-x^2$ meets the $x$-axis at $x=0,3$. By the shell method, the shell at $x$ has radius $x-(-1)=x+1$ and height $3x-x^2$, so $$V=2\pi\int_0^3 (x+1)(3x-x^2)\,dx = 2\pi\int_0^3(-x^3+2x^2+3x)\,dx$$ $$=2\pi\left[-\frac{x^4}{4}+\frac{2x^3}{3}+\frac{3x^2}{2}\right]_0^3=2\pi\cdot\frac{45}{4}=\frac{45}{2}\pi.$$

Solution. (1) Revolving about the $y$-axis, the shell at $x$ has radius $x$ and height $\sqrt x$, so $$V=2\pi\int_0^4 x\sqrt x\,dx = 2\pi\int_0^4 x^{3/2}\,dx=2\pi\left[\frac25x^{5/2}\right]_0^4=2\pi\cdot\frac25\cdot32=\frac{128}{5}\pi.$$

(2) Revolving about the $x$-axis, the shell at $y$ has radius $y$ and height $4-y^2$ (for $0\le y\le2$), so $$V=2\pi\int_0^2 y(4-y^2)\,dy=2\pi\left[2y^2-\frac{y^4}{4}\right]_0^2=2\pi(8-4)=8\pi.$$

Solution. Revolving about the $y$-axis, the shell at $x$ has radius $x$ and height $g(x)=\dfrac{\tan^2x}{x}$, so $$V=2\pi\int_0^{\pi/4} x\cdot\frac{\tan^2x}{x}\,dx=2\pi\int_0^{\pi/4}\tan^2x\,dx=2\pi\int_0^{\pi/4}(\sec^2x-1)\,dx$$ $$=2\pi\left[\tan x-x\right]_0^{\pi/4}=2\pi\left(1-\frac\pi4\right).$$

Solution. Revolving about the $x$-axis, the shell at $y$ has radius $y$ and height $$\frac{y^2}{2}-\left(\frac{y^4}{4}-\frac{y^2}{2}\right)=y^2-\frac{y^4}{4},$$ so $$V=2\pi\int_0^2 y\left(y^2-\frac{y^4}{4}\right)dy=2\pi\int_0^2\left(y^3-\frac{y^5}{4}\right)dy=2\pi\left[\frac{y^4}{4}-\frac{y^6}{24}\right]_0^2=2\pi\left(4-\frac83\right)=\frac83\pi.$$

Solution. (1) Since $y=\dfrac{x^3}{12}+\dfrac1x$, we have $y'=\dfrac{x^2}{4}-\dfrac{1}{x^2}$, and $$1+(y')^2 = 1+\left(\frac{x^2}{4}-\frac{1}{x^2}\right)^2=\frac{x^4}{16}+\frac12+\frac{1}{x^4}=\left(\frac{x^2}{4}+\frac{1}{x^2}\right)^2.$$ Since $x>0$, $\sqrt{1+(y')^2}=\dfrac{x^2}{4}+\dfrac{1}{x^2}$, so with base point $A=(1,13/12)$, $$s(x)=\int_1^x\left(\frac{t^2}{4}+\frac1{t^2}\right)dt=\left[\frac{t^3}{12}-\frac1t\right]_1^x=\frac{x^3}{12}-\frac1x-\left(\frac1{12}-1\right)=\frac{x^3}{12}-\frac1x+\frac{11}{12}.$$

(2) Using the integrand found in (1), $$L=\int_1^4\left(\frac{x^2}{4}+\frac1{x^2}\right)dx=\left[\frac{x^3}{12}-\frac1x\right]_1^4=\left(\frac{64}{12}-\frac14\right)-\left(\frac1{12}-1\right)=\frac{61}{12}+\frac{11}{12}=6.$$

Proof. Differentiating both sides of $9x^2=y(y-3)^2=y^3-6y^2+9y$ with respect to $y$ gives $$18x\frac{dx}{dy}=3y^2-12y+9=3(y-1)(y-3),$$ so $$\frac{dx}{dy}=\frac{(y-1)(y-3)}{6x}.$$ Since $36x^2=4y(y-3)^2$, $$\left(\frac{dx}{dy}\right)^2=\frac{(y-1)^2(y-3)^2}{36x^2}=\frac{(y-1)^2(y-3)^2}{4y(y-3)^2}=\frac{(y-1)^2}{4y}.$$ Therefore $$ds^2=\left[\left(\frac{dx}{dy}\right)^2+1\right]dy^2=\left(\frac{(y-1)^2}{4y}+1\right)dy^2=\frac{(y-1)^2+4y}{4y}\,dy^2=\frac{y^2+2y+1}{4y}\,dy^2=\frac{(y+1)^2}{4y}\,dy^2.$$

Solution. Since $y=2\sqrt x$, we have $y'=\dfrac{1}{\sqrt x}$, and $$1+(y')^2=1+\frac1x=\frac{x+1}{x}.$$ The area of the surface generated by revolving about the $x$-axis is $$S=2\pi\int_1^2 y\sqrt{1+(y')^2}\,dx=2\pi\int_1^2 2\sqrt x\cdot\sqrt{\frac{x+1}{x}}\,dx=2\pi\int_1^2 2\sqrt{x+1}\,dx$$ $$=4\pi\left[\frac23(x+1)^{3/2}\right]_1^2=\frac{8\pi}{3}\left(3\sqrt3-2\sqrt2\right).$$

Solution. (1) Since $x=1-y$, $\dfrac{dx}{dy}=-1$ and $\sqrt{1+(dx/dy)^2}=\sqrt2$. The area of the surface generated by revolving about the $y$-axis is $$S=2\pi\int_0^1 x\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy=2\pi\int_0^1(1-y)\sqrt2\,dy=2\sqrt2\pi\left[y-\frac{y^2}{2}\right]_0^1=2\sqrt2\pi\cdot\frac12=\sqrt2\pi.$$

(2) Since $y=\sqrt{2x-x^2}$, we have $y^2=2x-x^2=1-(x-1)^2$ and $y'=\dfrac{1-x}{\sqrt{2x-x^2}}$. Hence $$1+(y')^2=1+\frac{(x-1)^2}{2x-x^2}=\frac{2x-x^2+(x-1)^2}{2x-x^2}=\frac{1}{2x-x^2},$$ so $y\sqrt{1+(y')^2}=\sqrt{2x-x^2}\cdot\dfrac{1}{\sqrt{2x-x^2}}=1$. The area of the surface generated by revolving about the $x$-axis is $$S=2\pi\int_{1/2}^{3/2} 1\,dx=2\pi\left(\frac32-\frac12\right)=2\pi.$$

(3) Since $y=\dfrac13(x^2+2)^{3/2}$, we have $y'=x\sqrt{x^2+2}$, and $$1+(y')^2=1+x^2(x^2+2)=x^4+2x^2+1=(x^2+1)^2.$$ The area of the surface generated by revolving about the $y$-axis is $$S=2\pi\int_0^{\sqrt2} x\sqrt{1+(y')^2}\,dx=2\pi\int_0^{\sqrt2} x(x^2+1)\,dx=2\pi\left[\frac{x^4}{4}+\frac{x^2}{2}\right]_0^{\sqrt2}=2\pi(1+1)=4\pi.$$

Solution. (1) By Pappus's theorem, since the centroid of the boundary circle (radius $a$, circumference $2\pi a$) is at distance $b$ from the axis, the surface area of the torus is $$S=2\pi b\cdot 2\pi a=4\pi^2 ab.$$

(2) Parametrize the astroid as $x=\cos^3\theta$, $y=\sin^3\theta$ ($0\le\theta\le\pi$). Then $$\frac{dx}{d\theta}=-3\cos^2\theta\sin\theta,\qquad \frac{dy}{d\theta}=3\sin^2\theta\cos\theta,$$ so $$\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}=3|\cos\theta\sin\theta|.$$ By symmetry about the $y$-axis, doubling the contribution from $0\le\theta\le\pi/2$, the area of the surface generated by revolving about the $x$-axis is $$S=2\cdot2\pi\int_0^{\pi/2}\sin^3\theta\cdot 3\sin\theta\cos\theta\,d\theta=12\pi\int_0^{\pi/2}\sin^4\theta\cos\theta\,d\theta=12\pi\left[\frac{\sin^5\theta}{5}\right]_0^{\pi/2}=\frac{12}{5}\pi.$$