적분법 전체 문제

The intervals of our rectangles are $[0,\frac{\pi}{4}]$, $[\frac{\pi}{4},\frac{\pi}{2}]$, $[\frac{\pi}{2},\frac{3\pi}{4}]$, and $[\frac{3\pi}{4},\pi]$. Since we're taking a left Riemann sum, we find the height of the rectangles at the left endpoints of the intervals.

• $x = 0$: The distance from $\cos 0$ to $\sin 0$ is 1, so our first rectangle has height 1.
• $x = \frac{\pi}{4}$: The distance from $\cos \frac{\pi}{4}$ to $\sin \frac{\pi}{4}$ is 0, so our second rectangle has height 0.
• $x = \frac{\pi}{2}$: The distance from $\cos \frac{\pi}{2}$ to $\sin \frac{\pi}{2}$ is 1, so our third rectangle has height 1.
• $x = \frac{3\pi}{4}$: The distance from $\cos \frac{3\pi}{4}$ to $\sin \frac{3\pi}{4}$ is $\sin(3\pi/4)-\cos(3\pi/4) =\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)=\sqrt{2}$, so our fourth rectangle has height $\sqrt{2}$.

So, our approximation for the area between the two curves is $$\frac{\pi}{4}\left(1+0+1+\sqrt{2}\right)=\frac{\pi}{4}\left(2+\sqrt{2}\right)$$

(a) We are finding the area in the interval from $x=0$ to $x=\frac{\pi}{2}$. Since we're taking $n=5$ rectangles, our rectangles cover the following intervals: $$\left[0,\frac{\pi}{10}\right],\quad \left[\frac{\pi}{10},\frac{\pi}{5}\right],\quad \left[\frac{\pi}{5},\frac{3\pi}{10}\right],\quad \left[\frac{3\pi}{10},\frac{2\pi}{5}\right],\quad \left[\frac{2\pi}{5},\frac{\pi}{2}\right].$$

(b) We are finding the area in the interval from $y=0$ to $y=\frac{\pi}{2}$. (In general, when we switch from horizontal rectangles to vertical, the limits of integration will change–it's only coincidence that they are the same in this example.) Since we're taking $n=5$ rectangles, these rectangles cover the following intervals of the $y$-axis: $$\left[0,\frac{\pi}{10}\right],\quad \left[\frac{\pi}{10},\frac{\pi}{5}\right],\quad \left[\frac{\pi}{5},\frac{3\pi}{10}\right],\quad \left[\frac{3\pi}{10},\frac{2\pi}{5}\right],\quad \left[\frac{2\pi}{5},\frac{\pi}{2}\right].$$

The question doesn't specify which endpoints we're using. Let's use upper endpoints, to match part (a).

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 $\mathrm{d}y$.