적분법 전체 문제

We need to find where the curves intersect.

$$\begin{aligned} \frac{x^2}{4}=y^2&=6-\dfrac{5x}{4}\\ \frac{1}{4}x^2+\frac{5}{4}x-6&=0\\ x^2+5x-24&=0\\ (x+8)(x-3)&=0\\ x=-8,\quad x&=3 \end{aligned}$$

The curves intersect at $(-8,4)$ and $(3,-\frac{3}{2})$. Using horizontal strips:

we have

$$\begin{aligned} \text{Area} = \int_{-3/2}^{4}\Big[\frac{4}{5}(6-y^2)+2y\Big]\ \mathrm{d}y \end{aligned}$$

If the curves intersect at $(x,y)$, then

$$\begin{aligned} \left(x^2\right)^2&=\left(4a\right)^2y^2 = (4a)^24ax\\ x^4&=(4a)^3 x\\ x^4&-(4a)^3x=0\\ x(&x^3-(4a)^3)=0\\ x&= 0 \quad\text{or}\quad x^3=(4a)^3 \end{aligned}$$

The curves intersect at $(0,0)$ and $(4a,4a)$. (It is also possible to find these points by inspection.) Using vertical strips:

We want the $y$-values of the functions. We write the top function as $y =\sqrt{4ax}$ (we care about the positive square root, not the negative one) and we write the bottom function as $y=\frac{x^2}{4a}$. Then we have

$$\begin{aligned} \text{Area} = \int_0^{4a}\left[\sqrt{4ax}-\frac{x^2}{4a}\right]\ \mathrm{d}x \end{aligned}$$

The curves intersect when $x=4y^2$ and $0=4y^2+12y+5 =(2y+5)(2y+1)$. So, the curves intersect at $(1,-\tfrac{1}{2})$ and $(25,-\frac{5}{2})$. Using vertical strips:

we have

$$\begin{aligned} \text{Area} = \int_1^{25}\left[-\frac{1}{12}(x+5)+\frac{1}{2}\sqrt{x}\right]\ \mathrm{d}x \end{aligned}$$

(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

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