입체도형의 부피

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

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