적분법 전체 문제

17 kJ

If the curves $y=f(x)=x$ and $y=g(x)=3x-x^2$ intersect at $(x,y)$, then

$$\begin{aligned} 3x-x^2&=y=x\\ x^2-2x&=0\\ x(x-2)&=0\\ x=0 \quad &\text{or} \quad x=2 \end{aligned}$$

Furthermore, $g(x)-f(x) = 2x-x^2 = x(2-x)$ is positive for all $0\le x\le 2$. That is, the curve $y=3x-x^2$ lies above the line $y=x$ for all $0\le x\le 2$.

We therefore evaluate the integral:

$$\begin{aligned} \int_0^2 \big[ (3x-x^2) - x \big] \,\mathrm{d}x = \int_0^2 [2x-x^2]\,\mathrm{d}x = \bigg[x^2 - \frac{x^3}{3}\bigg]^{2}_{0} = \bigg[ 4-\frac{8}{3} \bigg] -0 = \frac{4}{3} \end{aligned}$$

For our computation, we will need an antiderivative of $x^2\sqrt{x^3+1}$, which can be found using the substitution $u=x^3+1$, $\mathrm{d}u = 3x^2\,\mathrm{d}x$:

$$\begin{aligned} \int x^2\sqrt{x^3+1} \, \mathrm{d}x = \int \sqrt u \cdot \frac13\,\mathrm{d}u = \frac13\int u^{1/2}\,\mathrm{d}u = \frac13\cdot \frac{u^{3/2}}{3/2}+C = \frac29(x^3+1)^{3/2} + C. \end{aligned}$$

The two functions $f(x)$ and $g(x)$ are clearly equal at $x=0$. If $x\ne0$, then the functions are equal when

$$\begin{aligned} 3x^2 &= x^2\sqrt{x^3+1} \\ 3 &= \sqrt{x^3+1} \\ 9 &= x^3+1 \\ 8 &= x^3 \\ 2 &= x. \end{aligned}$$

The function $g(x)=3x^2$ is the larger of the two on the interval $[0,2]$, as can be seen by plugging in $x=1$, say, or by observing that when $x$ is very small $f(x)=x^2\sqrt{x^3+1}\approx x^2$ and $g(x)=3x^2$.

The area in question is therefore:

$$\begin{aligned} \int_0^2 \big( 3x^2 - x^2\sqrt{x^3+1} \big) \, \mathrm{d}x &= \bigg( {x^3} - \frac29(x^3+1)^{3/2} \bigg) \bigg|_0^2 \\ &= \bigg( 2^3 - \frac2 9(2^3+1)^{3/2} \bigg) - \bigg( 0^3 - \frac29(0^3+1)^{3/2} \bigg) \\ &= \bigg( 8 - 6 \bigg) - \bigg( 0 - \frac 2 9 \bigg) =\frac{20}9. \end{aligned}$$

Let's begin by sketching our region. Note that $y=\sqrt{1-x^2}$ and $y=\sqrt{9-x^2}$ are the top halves of circles centred at the origin with radii 1 and 3, respectively.

Our region is the difference of two quarter-circles, so we find its area using geometry: $$\text{Area}=\frac{1}{4}\left(\pi\cdot 3^2\right)-\frac{1}{4}\left(\pi\cdot 1^2\right)=2\pi$$

The curves $y=1/x$ and $3x+3y=10$, i.e. $y =\frac{10}{3}-x$ intersect when

$$\begin{aligned} \frac{1}{x} = \frac{10}{3}-x &\iff 3 = 10x-3x^2 \iff 3x^2-10x+3=0 \\ &\iff(3x-1)(x-3)=0 \\ &\iff x=3\,,\,\frac{1}{3} \end{aligned}$$

enterline

When the region is rotated about the $x$–axis, the vertical strip in the figure above sweeps out a washer with thickness $\mathrm{d}x$, outer radius $T(x)=\frac{10}{3}-x$ and inner radius $B(x)=\frac{1}{x}$. This washer has volume

$$\pi\big(T(x)^2- B(x)^2\big)\,\mathrm{d}x = \pi\Big(\frac{100}{9}-\frac{20}{3}x+x^2-\frac{1}{x^2}\Big)\,\mathrm{d}x$$

Hence the volume of the solid is

$$\begin{aligned} \pi\int_{1/3}^3\Big(\frac{100}{9}-\frac{20}{3}x+x^2-\frac{1}{x^2}\Big)\,\mathrm{d}x &=\pi\Big[\frac{100x}{9}-\frac{10}{3}x^2+\frac{1}{3}x^3 +\frac{1}{x}\Big]_{1/3}^3 \\ &=\pi\Big[\frac{38}{3}-\frac{514}{3^4}\Big] = \pi\frac{512}{81} \end{aligned}$$

(a) The top and the bottom of the circle have equations $y=T(x)=2+\sqrt{1-x^2}$ and $y=B(x) = 2-\sqrt{1-x^2}$, respectively.

enterline

When $R$ is rotated about the $x$–axis, the vertical strip of $R$ in the figure above sweeps out a washer with thickness $\mathrm{d}x$, outer radius $T(x)$ and inner radius $B(x)$. This washer has volume

$$\pi\big(T(x)^2- B(x)^2\big)\,\mathrm{d}x = \pi\big(T(x)+ B(x)\big)\big(T(x)- B(x)\big)\,\mathrm{d}x = \pi\times 4\times 2\sqrt{1-x^2}\,\mathrm{d}x$$

Hence the volume of the solid is

$$8\pi\int_{-1}^1\sqrt{1-x^2}\,\mathrm{d}x$$

(b) Since $y=\sqrt{1-x^2}$ is equivalent to $x^2+y^2=1$, $y\ge 0$, the integral is $8\pi$ times the area of the upper half of the circle $x^2+y^2=1$ and hence is $8\pi\times \frac{1}{2}\pi 1^2 = 4\pi^2$.

Here is a sketch of the curves $y = \cos(\frac x2)$ and $y = x^2 - \pi^2$.

By inspection, the curves meet at $x = \pm {\pi}$ where both $\cos(\frac x2)$ and $x^2 - \pi^2$ take the value zero. We'll use vertical washers as specified in the question.

• We cut the specified region into thin vertical strips of width $\mathrm{d}x$ as in the figure above.

• When we rotate about the line $y=-\pi^2$, each strip sweeps out a thin washer

• whose inner radius is $r_{in}= (x^2 - {\pi^2} ) - ( {-} {\pi^2} )=x^2$ and outer radius is $r_{out}=\cos(\frac x2) - ( {-} {\pi^2}) =\cos(\frac x2) +\pi^2$, and

• whose thickness is $\mathrm{d}x$ and hence

• whose volume $\pi(r_{out}^2 - r_{in}^2)\mathrm{d}x = \pi\big( {(\cos(\frac x2) +\pi^2)}^2 - {(x^2)}^2\big)\mathrm{d}x$.

• As our leftmost strip is at $x=-\pi$ and our rightmost strip is at $x=\pi$,

the total volume is

$$\begin{aligned} &\pi \int_{-\pi}^{\pi} \left( \cos^2 (\tfrac x2) +2{\pi^2}\cos (\tfrac x2) +{\pi^4} -x^4\right)\,\mathrm{d}x \\ & = \pi \int_{-\pi}^{\pi} \left(\frac{1+\cos(x)}{2} +2{\pi^2}\cos (\tfrac x2) +{\pi^4} -x^4\right)\,\mathrm{d}x \\ \text{Because the integrand is even,} \\ & = 2\pi \int_0^{\pi} \left(\frac{1+\cos(x)}{2} +2{\pi^2}\cos (\tfrac x2) +{\pi^4} -x^4\right)\,\mathrm{d}x \\ & = {2\pi \left[ \frac{1}{2}x + \frac{1}{2}\sin(x) + 4{\pi^2}\sin (\tfrac x2) +{\pi^4} x -\frac{1}{5}x^5 \right]} _0^\pi\\ & = 2\pi \left[\frac{\pi}{2} + 0 + 4{\pi^2} +{\pi^5} - \frac{\pi^5}{5} \right] \\ & = {\pi^2} + 8\pi^3 + \frac{8\pi^6}{5} \end{aligned}$$

We used the fact that the integrand is an even function and the interval of integration $[-\pi, \pi]$ is symmetric, but one can also compute directly.