적분의 활용 전체 문제

Let's start by converting from time spent pulling to amount pulled. When $y$ metres of rope have been pulled up, $2y$ seconds have passed, so $\frac{1}{5}y$ litres of water have leaked out of the bucket, leaving $5-\frac{1}{5}y$ litres. (This only makes sense when $\frac{1}{5}y \leq 5$, but we only consider values of $y$ from 0 to 5, so it's not a problem. That is, we're never hauling up an empty bucket that can't leak any more.)

When we've pulled up $y$ metres of rope, the mass in the bucket is $(5-\frac{1}{5}y)$ kg, so the force of gravity acting on it is $9.8(5-\frac{1}{5}y)$ N. Since we pull up 5 metres of rope, the work done is: $$W = \int_0^5 9.8\left(5-\frac{1}{5}y\right) \,\mathrm{d}y = 9.8\left[5y-\frac{1}{10}y^2\right]_0^5 = 9.8\left[25-2.5\right]=220.5\text{ J}$$

(a) Here is a sketch of the specified region, which we shall call $R$.

The top of $R$ has equation $y=T(x)=e^x$, the bottom has equation $y=B(x)=-1$ and $x$ runs from $0$ to $1$. So, using vertical strips, we see that $R$ has area

$$\begin{aligned} A & = \int_0^1 \big[T(x)-B(x)\big]\,\mathrm{d}x = \int_0^1 \big[e^x-(-1)\big]\,\mathrm{d}x = \int_0^1 \big[e^x+1\big]\,\mathrm{d}x = \big[e^x+x\big]_0^1 = e \end{aligned}$$

and

$$\begin{aligned} \bar y &=\frac{1}{2A} \int_0^1 \big[T(x)^2-B(x)^2\big]\,\mathrm{d}x \\ &= \frac{1}{2e}\int_0^1 \big[ e^{2x} - 1\ \big] \,\mathrm{d}x = \frac{1}{2e} \bigg[\frac{e^{2x}}{2} - x \bigg]_0^1 \\ &=\frac{1}{2e} \bigg(\frac{e^2}{2} - 1 - \frac{1}{2} \bigg) = \frac{e}{4} - \frac{3}{4e} \end{aligned}$$

(b) To compute the volume when $R$ is rotated about the line $y=-1$

• we use thin vertical strips of width $\mathrm{d}x$ as in the figure above.

• When we rotate about the line $y=-1$, each strip sweeps out a thin disk

• whose radius is $r=T(x)-B(x)=e^x+1$ and

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

• whose volume is $\pi r^2\,\mathrm{d}x = \pi(e^x+1)^2\,\mathrm{d}x$.

• As our leftmost strip is at $x=0$ and our rightmost strip is at $x=1$, the total volume is

$$\begin{aligned} \pi \int_0^1 (e^x+1)^2\,\mathrm{d}x &=\pi \int_0^1 (e^{2x}+2e^x+1)\,\mathrm{d}x =\pi\bigg[\frac{e^{2x}}{2}+2e^x + x\bigg]_0^1 \\ &=\pi\bigg[\Big(\frac{e^{2}}{2}+2e + 1\Big) -\Big(\frac{1}{2}+2 + 0\Big)\bigg] \\ &=\pi\Big(\frac{e^{2}}{2}+2e -\frac{3}{2}\Big) \end{aligned}$$