Unit B4: Applications of Integration: Area, Volume, Arc LengthB4 单元：积分的应用——面积、体积、弧长

Find the area enclosed by $y=x^2$ and $y=x+2$.

First locate the intersections by setting $x^2=x+2$, so $x^2-x-2=(x-2)(x+1)=0$, giving $x=-1$ and $x=2$. On $(-1,2)$ test $x=0$: the line gives $2$ and the parabola gives $0$, so the line is on top.

At $x=2$ the bracket is $2+4-\tfrac{8}{3}=\tfrac{10}{3}$; at $x=-1$ it is $\tfrac12-2+\tfrac13=-\tfrac{7}{6}$. Subtracting,

求 $y=x^2$ 与 $y=x+2$ 围成的面积。

先求交点：令 $x^2=x+2$，即 $x^2-x-2=(x-2)(x+1)=0$，得 $x=-1$ 和 $x=2$。在 $(-1,2)$ 上取 $x=0$ 验证：直线值为 $2$，抛物线值为 $0$，故直线在上方。

在 $x=2$ 处括号值为 $\tfrac{10}{3}$；在 $x=-1$ 处为 $-\tfrac{7}{6}$。作差得

Find the area bounded by $x=y^2$ and $x=y+2$.

Intersections: $y^2=y+2$ gives $y^2-y-2=(y-2)(y+1)=0$, so $y=-1$ and $y=2$. For $-1 $$A=\int_{-1}^{2}\bigl((y+2)-y^2\bigr)\,dy=\frac{9}{2},$$

by the same antiderivative as Example 1.1. Slicing in $y$ avoids splitting the region that a vertical slice would require, since the left boundary of the region would otherwise be the two branches of the sideways parabola.

求 $x=y^2$ 与 $x=y+2$ 围成的面积。

求交点：$y^2=y+2$ 即 $(y-2)(y+1)=0$，得 $y=-1$ 和 $y=2$。在 $-1 $$A=\int_{-1}^{2}\bigl((y+2)-y^2\bigr)\,dy=\frac{9}{2},$$

原函数（antiderivative）与例题 1.1 相同。关于 $y$ 切片避免了竖直切片所需的区间拆分，否则左边界将是横向抛物线的两支。

Find the geometric area between $y=\sin x$ and $y=\cos x$ on $[0,\pi/2]$.

These curves cross where $\sin x=\cos x$, that is at $x=\pi/4$. On $[0,\pi/4)$ we have $\cos x>\sin x$ (test $x=0$: $1>0$); on $(\pi/4,\pi/2]$ the order reverses (test $x=\pi/2$: $1>0$ for $\sin$). Because the top curve changes, a single integrand $(\cos x-\sin x)$ over the whole interval would record a signed area in which the second piece subtracts. Split at the crossing and take top minus bottom on each piece:

The first integral is $[\sin x+\cos x]_0^{\pi/4}=\bigl(\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}\bigr)-(0+1)=\sqrt2-1$. The second is $[-\cos x-\sin x]_{\pi/4}^{\pi/2}=(0-1)-\bigl(-\tfrac{\sqrt2}{2}-\tfrac{\sqrt2}{2}\bigr)=\sqrt2-1$. By symmetry the two pieces are equal, so

Suppose first that both curves lie above the $x$-axis, $f(x)\ge g(x)\ge 0$. The region between them is the region under $f$ with the region under $g$ removed. By the area-under-a-curve interpretation of the definite integral,

using linearity of the integral. Now drop the sign assumption. Choose a constant $C$ large enough that $g(x)+C\ge 0$ on $[a,b]$ (possible since $g$ is continuous on a closed interval, hence bounded below). Shifting both curves up by $C$ is a vertical translation that preserves the enclosed area, and the shifted curves satisfy the nonnegative hypothesis, so

The constant cancels, which is why the formula needs no positivity assumption: only the ordering $f\ge g$ matters, and that is what forces the integrand to be the gap, top minus bottom.

Rotate the region under $y=\sqrt{r^2-x^2}$, $-r\le x\le r$, about the $x$-axis to recover the volume of a sphere of radius $r$.

The region between $y=x$ and $y=x^2$ on $[0,1]$ is revolved about the $x$-axis. Find the volume.

On $(0,1)$ the line $y=x$ is the outer boundary and $y=x^2$ the inner, so $R(x)=x$ and $r(x)=x^2$.

The region bounded by $y=x^2$ and $y=1$ is revolved about the horizontal line $y=2$. Find the volume.

The region runs over $-1\le x\le 1$, with $y$ between the parabola $y=x^2$ (bottom) and the line $y=1$ (top). Revolving about $y=2$ produces a washer at each $x$. A radius is the distance from the axis $y=2$ down to a boundary curve. The point on the region closest to the axis is the top edge $y=1$, at distance $2-1=1$: that is the inner radius. The farthest is the parabola $y=x^2$, at distance $2-x^2$: that is the outer radius.

The integrand is even, so integrate on $[0,1]$ and double:

The region under $y=2x-x^2$ on $[0,2]$ is revolved about the $y$-axis. Find the volume.

A vertical strip at $x$ has radius $x$ and height $2x-x^2$.

Consider the shell between radius $x$ and $x+dx$, with height $h$. Its exact volume is the difference of two cylinders,

Divide by $dx$ and let $dx\to 0$. The term $(dx)^2$ is second order and vanishes, leaving the density $dV/dx=2\pi x h$. Equivalently, unroll the shell into a thin rectangular sheet of length $2\pi x$ (the circumference), height $h$, and thickness $dx$; its volume is $2\pi x\cdot h\cdot dx$. Summing over all strips gives the shell integral.

The region under $y=x^2$ on $[0,1]$ is revolved about the line $x=2$. Find the volume by shells.

A vertical strip sits at position $x$ with height $x^2$. Revolving about $x=2$, the strip is a distance $2-x$ from the axis (the region lies to the left of the line, so the radius is $2-x$, which is positive on $[0,1]$). The shell height is still $f(x)=x^2$.

Revolve the region bounded by $y=x^3$, $y=0$, and $x=2$ about the $y$-axis.

A strip at $x$ has radius $x$ and runs from the $x$-axis up to $y=x^3$, so its height is $x^3$.

Washers in $y$ would require the inverse $x=y^{1/3}$ and an outer radius of $2$, a workable but longer computation; shells read directly off the original equation.

Revolve the region under $y=x^2$ on $[0,3]$ about the $y$-axis. We found $V=\tfrac{81\pi}{2}$ by shells in Section 3. Confirm by washers in $y$.

Slicing perpendicular to the $y$-axis, at height $y$ the solid runs from the curve $x=\sqrt{y}$ out to the line $x=3$, so a washer has outer radius $3$ and inner radius $\sqrt{y}$, for $0\le y\le 9$.

The two methods agree, as they must.

Revolve the region under $y=\sin x$ on $[0,\pi]$ about the $y$-axis.

Washers in $y$ would require inverting $y=\sin x$ into two arcs, an unpleasant split. Shells in $x$ are immediate:

Integrate by parts with $u=x$, $dv=\sin x\,dx$: $\int x\sin x\,dx=-x\cos x+\sin x$. Evaluating from $0$ to $\pi$ gives $-\pi\cos\pi+\sin\pi-0=\pi$. Hence $V=2\pi\cdot\pi=2\pi^2$.

Revolve the region bounded by $y=\sqrt{x}$, $y=0$, and $x=4$ about the $x$-axis. Compute by disks, then verify by shells.

Disks (slices perpendicular to the $x$-axis). At $x$ the radius is $\sqrt{x}$:

Shells (slices parallel to the $x$-axis, i.e. horizontal strips in $y$). A horizontal strip at height $y$ runs in $x$ from the curve $x=y^2$ to the line $x=4$, so its length is $4-y^2$, and its distance to the $x$-axis is $y$, for $0\le y\le 2$:

The methods agree at $8\pi$, which is the cross-check you want before trusting a single setup.

Find the length of $y=\tfrac{x^3}{6}+\tfrac{1}{2x}$ on $[1,2]$.

Differentiate: $f'(x)=\tfrac{x^2}{2}-\tfrac{1}{2x^2}$. Then

So the square root is $\tfrac{x^2}{2}+\tfrac{1}{2x^2}$, and

Find the length of the curve $x=\tfrac{1}{3}(y^2+2)^{3/2}$ from $y=0$ to $y=3$.

This curve is naturally a function of $y$, so use $L=\int_c^d\sqrt{1+(g'(y))^2}\,dy$. Differentiate:

Then $(g')^2=y^2(y^2+2)=y^4+2y^2$, so

a perfect square as designed. The integrand is $\sqrt{(y^2+1)^2}=y^2+1$ (nonnegative on $[0,3]$), giving

Set up the length of $y=x^2$ from $x=0$ to $x=1$.

Here $f'(x)=2x$, so $1+(f')^2=1+4x^2$, which is not a perfect square. The exact length is

This does have a closed form via the substitution $2x=\tan\theta$ or the standard reduction, namely $\tfrac{1}{4}\bigl[2x\sqrt{1+4x^2}+\sinh^{-1}(2x)\bigr]_0^1=\tfrac{1}{4}\bigl(2\sqrt5+\sinh^{-1}2\bigr)\approx 1.479$. The lesson is procedural: most arc length integrands have no elementary perfect-square form, so on an exam you typically stop at the correct setup unless the problem is engineered, then evaluate numerically only if asked.

Partition $[a,b]$ as $a=x_0 $$L_n=\sum_{i=1}^n \sqrt{(\Delta x_i)^2+(\Delta y_i)^2}=\sum_{i=1}^n \sqrt{1+\left(\frac{\Delta y_i}{\Delta x_i}\right)^2}\,\Delta x_i.$$

Because $f$ is differentiable, the Mean Value Theorem gives a point $x_i^*$ in each subinterval with $\Delta y_i/\Delta x_i=f'(x_i^*)$. Thus $L_n=\sum_i \sqrt{1+f'(x_i^*)^2}\,\Delta x_i$, a Riemann sum for $\sqrt{1+(f')^2}$. As the mesh tends to zero this converges to $\int_a^b\sqrt{1+(f')^2}\,dx$, provided $f'$ is continuous so the integrand is integrable.

One subtlety justifies calling $L$ the length: the polygonal lengths $L_n$ increase as the partition is refined (adding a vertex can only lengthen the path, by the triangle inequality), and the arc length is defined as the supremum of all polygonal lengths. The Riemann-sum limit above realizes that supremum when $f'$ is continuous, so the integral and the geometric definition coincide.

Revolve $y=\sqrt{x}$ on $[0,1]$ about the $x$-axis and find the surface area.

Here $f(x)=x^{1/2}$, $f'(x)=\tfrac{1}{2}x^{-1/2}$, so $1+(f')^2=1+\tfrac{1}{4x}=\tfrac{4x+1}{4x}$. Then

With $u=4x+1$, $du=4\,dx$, this is $\pi\cdot\tfrac14\cdot\tfrac23 u^{3/2}\big|_{1}^{5}=\dfrac{\pi}{6}\bigl(5^{3/2}-1\bigr)=\dfrac{\pi}{6}\bigl(5\sqrt5-1\bigr).$

Revolve the line $y=\tfrac{r}{h}x$ on $[0,h]$ about the $x$-axis to recover the lateral surface area of a cone of base radius $r$ and height $h$.

Here $f'=\tfrac{r}{h}$ is constant, so $\sqrt{1+(f')^2}=\sqrt{1+r^2/h^2}=\tfrac{\sqrt{h^2+r^2}}{h}$, the reciprocal of $h$ times the slant length $\ell=\sqrt{h^2+r^2}$. Then

the familiar formula $\pi r\ell$ for the lateral area of a cone.

Revolve the upper semicircle $y=\sqrt{r^2-x^2}$, $-r\le x\le r$, about the $x$-axis to recover the surface area of a sphere of radius $r$.

Differentiate: $f'(x)=\dfrac{-x}{\sqrt{r^2-x^2}}$, so $(f')^2=\dfrac{x^2}{r^2-x^2}$ and

Now the radius $f(x)=\sqrt{r^2-x^2}$ multiplies the slant element, and a beautiful cancellation occurs:

The integrand collapses to the constant $2\pi r$, which is Archimedes' striking observation that equal slabs of a sphere carry equal surface area.

Over a subinterval the curve is approximated by a straight chord, and revolving that chord sweeps a frustum (a cone with the tip cut off), not a flat ring. A frustum with smaller radius $r_1$, larger radius $r_2$, and slant height $\ell$ has lateral area

The factor $\tfrac{r_1+r_2}{2}$ is the average radius, which in the limit becomes $f(x_i^*)$ at a sample point, and the slant height of the chord is $\ell=\sqrt{(\Delta x)^2+(\Delta y)^2}=\sqrt{1+(f')^2}\,\Delta x$. Therefore each frustum contributes $2\pi f(x_i^*)\sqrt{1+(f'(x_i^*))^2}\,\Delta x$, and summing then taking the limit gives the surface area integral. This is exactly why the integrand uses the slant element $ds$ and not the flat width $dx$: a tilted band has more surface than its horizontal shadow.

A solid has base the disk $x^2+y^2\le 1$ and cross sections perpendicular to the $x$-axis that are squares. Find its volume.

At position $x$ the base chord runs from $y=-\sqrt{1-x^2}$ to $y=\sqrt{1-x^2}$, a length $s(x)=2\sqrt{1-x^2}$. The square cross section has area $A(x)=s(x)^2=4(1-x^2)$.

Let a region $D$ in the right half plane be revolved about the $y$-axis. By shells, $V=\int 2\pi x\,dA$, where $dA$ is the area element and $x$ its distance to the axis. Factor out the constant $2\pi$:

The middle factor is by definition the centroidal coordinate $\bar{x}$ and the last is the area $A$. Hence $V=2\pi\bar{x}A$, which is Pappus's theorem: the volume equals the area times the distance the centroid travels, $2\pi\bar{x}$.

A solid has base the region bounded by $y=x$, $y=0$, and $x=1$. Cross sections perpendicular to the $x$-axis are equilateral triangles standing on the base. Find the volume.

At position $x$ the base region runs in $y$ from $0$ to $x$, so the side of the triangle is $s(x)=x$. An equilateral triangle of side $s$ has area $\tfrac{\sqrt3}{4}s^2$, so $A(x)=\tfrac{\sqrt3}{4}x^2$.

No revolution enters; the only ingredient is a correct area formula for the cross section as a function of $x$.

Pappus's second theorem states that the surface area generated by revolving a plane curve about an external axis is $S=2\pi\bar{r}\,L$, where $L$ is the curve length and $\bar{r}$ the distance from the curve's centroid to the axis. Apply it to the circle of radius $1$ centered at $(3,0)$ revolved about the $y$-axis, which generates a torus.

The generating curve is the circle itself, of length $L=2\pi(1)=2\pi$, and its centroid is the center $(3,0)$, so $\bar{r}=3$. Then

Compare with the volume of the same torus, $V=2\pi\bar{r}A=2\pi(3)(\pi)=6\pi^2$, from the unit quiz. The two Pappus theorems handle volume (using the region's area) and surface area (using the boundary curve's length) by the same centroid-travel idea.