입체도형의 부피

$175\pi/3$ units$^3$

$\pi^2/4$ units$^3$

$\pi/6$ units$^3$

$9\pi/2$ units$^3$

$35\pi/3$ units$^3$

$\pi^2-2\pi$ units$^3$

$2\pi/15$ units$^3$

(a) $\pi/2$ (b) $5\pi/6$ (c) $4\pi/5$ (d) $8\pi/15$

(a) $512\pi/15$ (b) $256\pi/5$ (c) $832\pi/15$ (d) $128\pi/3$

(a) $4\pi/3$ (b) $2\pi/3$ (c) $4\pi/3$ (d) $\pi/3$

(a) $104\pi/15$ (b) $64\pi/15$ (c) $32\pi/5$

(a) $\pi^2/2$ (b) $\pi^2/2-4\pi\sinh^{-1} (1)$ (c) $\pi^2/2+4\pi\sinh^{-1} (1)$

(a) $8\pi$ (b) $8\pi$ (c) $16\pi/3$ (d) $8\pi/3$

The columns have the same volume. We can see this by chopping up the columns into horizontal cross-sections. Each cross-section has the same area as the cookie cutter, $A$, and height $\mathrm{d}y$. Then in both cases, the volume of the column is $$\int_{0}^h A~\mathrm{d}y = hA \text{ cubic units}$$

Notice $f(x)$ is a piecewise linear function, so we can find explicit equations for each of its pieces from the graph. The radii will be determined by the $x$-values, so below we give the $x$-values as functions of $y$.

If we imagine rotating the region from the picture about the $y$-axis, there will be two kinds of washers formed: when $y<1$, we have a “double washer," two concentric rings. When $y>1$, we have a single ring.

• Washers when $\mathbf{1<y \le 6}$: If $y>1$, then our washer has inner radius $2+\frac{2}{3}y$, outer radius $6-\frac{2}{3}y$, and height $\mathrm{d}y$.

• Washers when $\mathbf{0\le y <1}$: When $0 \le y < 1$, we have a “double washer," two concentric rings corresponding to the two “humps" in the function. The inner washer has inner radius $r_1=y$ and outer radius $R_1=2-y$. The outer washer has inner radius $r_2=2+\frac{2}{3}y$ and outer radius $R_2=6-\frac{2}{3}y$. The thickness of the washers is $\mathrm{d}y$.

(a) The curves intersect at $(1,0)$ and $(-1,0)$. When the strip shown in the figure

is rotated about the line $y=-1$, it forms a thin washer with:

• inner radius $(1-x^2)-(-1)=2-x^2$,
• outer radius $(4-4x^2)-(-1)=5-4x^2$ and
• thickness $\mathrm{d}x$ ; so, it has
• cross sectional area $\pi\big[{(5-4x^2)}^2-{(2-x^2)}^2\big]$ and
• volume $\pi\big[{(5-4x^2)}^2-({2-x^2)}^2\big]\,\mathrm{d}x$.

So the volume of the solid is

$$\begin{aligned} \int_{-1}^{1}\pi\big[{(5-4x^2)}^2-{(2-x^2)}^2\big]\,\mathrm{d}x \end{aligned}$$

(b) The curve $y=x^2-1$ intersects $y=0$ at $(1,0)$ and $(-1,0)$.

We'll use horizontal washers.

• We use thin horizontal strips of height $\mathrm{d}y$ as in the figure above.

• When we rotate about the line $x=5$, each strip sweeps out a thin washer

• whose inner radius is $r_{in}=5-\sqrt{y+1}$, and

• whose outer radius is $r_{out}= 5-(-\sqrt{y+1})=5+\sqrt{y+1}$ and

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

• whose volume is $\pi(r_{out}^2 - r_{in}^2)\,\mathrm{d}y = \pi\big[\big(5+\sqrt{y+1}\big)^2-\big(5-\sqrt{y+1}\big)^2\big]\,\mathrm{d}y$

• As our topmost strip is at $y=0$ and our bottommost strip is at $y=-1$ (when $x=0$), the total volume is

$$\begin{aligned} \int _{-1}^0 \pi\left[\big(5+\sqrt{y+1}\big)^2-\big(5-\sqrt{y+1}\big)^2\right]\,\mathrm{d}y \end{aligned}$$

The curves intersect at $(-2,4)$ and $(2,4)$. When the strip shown in the figure

is rotated about the line $y=-1$, it forms a thin washer (punctured disc) of

• inner radius $x^2+1$,
• outer radius $9-x^2$ and
• thickness $\mathrm{d}x$ and hence of
• cross sectional area $\pi\big[{(9-x^2)}^2-{(x^2+1)}^2\big]$ and
• volume $\pi\big[{(9-x^2)}^2-{(x^2+1)}^2\big]\,\mathrm{d}x$.

So the volume of the solid is

$$\begin{aligned} \pi\int_{-2}^{2}\big[{(9-x^2)}^2-{(x^2+1)}^2\big]\ \mathrm{d}x \end{aligned}$$

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.