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Chapter 8: Applications of Integration

AP Calculus AB/BC · Updated February 2026 · 20 min read · AP Calculus Past Papers

Integration is far more than an algebraic exercise in finding antiderivatives. Its real power lies in the geometric and physical problems it can solve. In this chapter we use definite integrals to compute areas between curves, volumes of three-dimensional solids, arc lengths, and average values of functions. These topics appear every year on the AP Calculus exam, and volumes of revolution in particular are among the most heavily tested concepts on both the AB and BC exams.

Table of Contents

  1. Area Between Curves
  2. Volumes of Revolution: Disk and Washer Method
  3. Volumes of Revolution: Shell Method BC
  4. Volumes with Known Cross Sections
  5. Arc Length BC
  6. Average Value of a Function
  7. Practice Problems

8.1 Area Between Curves

When two continuous functions $f(x)$ and $g(x)$ are defined on the interval $[a, b]$ and $f(x) \ge g(x)$ throughout that interval, the region trapped between their graphs has a well-defined area. We compute it by slicing the region into infinitely thin vertical rectangles, each of height $f(x) - g(x)$ and width $dx$, then summing them up.

Definition: Area Between Two Curves (Vertical Slices)
If $f(x) \ge g(x)$ on $[a, b]$, the area of the region between the curves is $$A = \int_a^b \bigl[f(x) - g(x)\bigr]\,dx$$ The rule is always top minus bottom. If the curves cross, you must split the integral at each intersection point and take the absolute value of each piece (or equivalently, always subtract the lower curve from the upper one on each subinterval).

Sometimes the region is easier to describe using horizontal slices. In that case, we express the boundary curves as functions of $y$ and integrate with respect to $y$.

Definition: Area Between Two Curves (Horizontal Slices)
If $x = h(y)$ is to the right of $x = k(y)$ on $[c, d]$, then $$A = \int_c^d \bigl[h(y) - k(y)\bigr]\,dy$$ The rule becomes right minus left.
Example 8.1.1: Find the area between $f(x) = x + 2$ and $g(x) = x^2$ from $x = -1$ to $x = 2$.

Solution: First check which function is on top. For $x$ in $[-1, 2]$, we have $f(x) \ge g(x)$ (you can verify at a sample point like $x = 0$: $f(0) = 2 > 0 = g(0)$). So: $$A = \int_{-1}^{2}\bigl[(x+2) - x^2\bigr]\,dx = \int_{-1}^{2}\bigl(-x^2 + x + 2\bigr)\,dx$$ $$= \left[-\frac{x^3}{3} + \frac{x^2}{2} + 2x\right]_{-1}^{2}$$ At $x = 2$: $-\frac{8}{3} + 2 + 4 = -\frac{8}{3} + 6 = \frac{10}{3}$.
At $x = -1$: $\frac{1}{3} + \frac{1}{2} - 2 = \frac{2 + 3 - 12}{6} = -\frac{7}{6}$.
$$A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{10}{3} + \frac{7}{6} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$$
Example 8.1.2: Find the area enclosed by $y = x^2$ and $y = 2x$.

Solution: First find the intersection points. Set $x^2 = 2x$, so $x^2 - 2x = 0$, giving $x(x-2) = 0$. The curves intersect at $x = 0$ and $x = 2$. Between these values, $2x \ge x^2$ (check $x = 1$: $2 > 1$). Therefore: $$A = \int_0^2 (2x - x^2)\,dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{12 - 8}{3} = \frac{4}{3}$$
Example 8.1.3 (Using $dy$): Find the area enclosed by $x = y^2$ and $x = y + 2$.

Solution: Find intersections: set $y^2 = y + 2$, so $y^2 - y - 2 = 0$, giving $(y-2)(y+1) = 0$. The curves meet at $y = -1$ and $y = 2$. For $y$ in $[-1, 2]$, the line $x = y + 2$ is to the right of the parabola $x = y^2$ (check $y = 0$: $2 > 0$). Using horizontal slices (right minus left): $$A = \int_{-1}^{2}\bigl[(y + 2) - y^2\bigr]\,dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^{2}$$ At $y = 2$: $2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}$.
At $y = -1$: $\frac{1}{2} - 2 + \frac{1}{3} = -\frac{7}{6}$.
$$A = \frac{10}{3} + \frac{7}{6} = \frac{20 + 7}{6} = \frac{27}{6} = \frac{9}{2}$$
AP Tip: If the boundary curves are given as functions of $y$, or if the region requires splitting into multiple integrals in $x$ but only one integral in $y$, it is almost always easier to integrate with respect to $y$.

Area between curves: drag the sliders to adjust the bounds a and b. The shaded region shows ∫[f(x)−g(x)] dx where f(x)=x+2 (blue) and g(x)=x² (red).

Area between curves: shaded region = ∫−12 [(x+2) − x²] dx = 9/2

8.2 Volumes of Revolution: Disk and Washer Method

When a region in the plane is revolved around an axis, it sweeps out a three-dimensional solid of revolution. The disk/washer method computes the volume of such a solid by slicing it into thin circular disks (or annular washers) perpendicular to the axis of rotation, then integrating the cross-sectional areas.

Disk Method
When a region bounded by $y = f(x)$ and the $x$-axis is revolved about the $x$-axis, each cross section is a disk of radius $R(x) = f(x)$: $$V = \pi \int_a^b \bigl[R(x)\bigr]^2\,dx$$
Washer Method
When the region between $y = f(x)$ (outer) and $y = g(x)$ (inner) is revolved about the $x$-axis, each cross section is an annular ring with outer radius $R(x) = f(x)$ and inner radius $r(x) = g(x)$: $$V = \pi \int_a^b \Bigl(\bigl[R(x)\bigr]^2 - \bigl[r(x)\bigr]^2\Bigr)\,dx$$

The key to setting up these integrals correctly is to identify two things: (1) the axis of rotation, and (2) the distance from each boundary curve to that axis. When revolving around a horizontal axis, you integrate with respect to $x$. When revolving around a vertical axis using disk/washer, you integrate with respect to $y$.

AP Tip: The radii are always measured from the axis of rotation, not from the $x$-axis. When the axis is $y = k$ instead of $y = 0$, the outer radius becomes $R(x) = f(x) - k$ (or $k - f(x)$ if the curve is below the axis). Drawing a sketch with one representative slice is the single most important step.
Example 8.2.1 (Disk Method): Find the volume of the solid formed by revolving $y = \sqrt{x}$ about the $x$-axis, from $x = 0$ to $x = 4$.

Solution: Each cross section perpendicular to the $x$-axis is a disk of radius $R(x) = \sqrt{x}$. $$V = \pi \int_0^4 \bigl(\sqrt{x}\bigr)^2\,dx = \pi \int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi \cdot \frac{16}{2} = 8\pi$$
Example 8.2.2 (Washer Method): Find the volume of the solid formed by revolving the region between $y = x^2$ and $y = x$ about the $x$-axis.

Solution: The curves intersect where $x^2 = x$, i.e., at $x = 0$ and $x = 1$. On $[0, 1]$, $y = x \ge y = x^2$, so when revolved about the $x$-axis, $R(x) = x$ (outer) and $r(x) = x^2$ (inner). $$V = \pi \int_0^1 \bigl(x^2 - x^4\bigr)\,dx = \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \pi \cdot \frac{2}{15} = \frac{2\pi}{15}$$
Example 8.2.3 (Revolving Around the $y$-axis): Find the volume of the solid formed by revolving $y = x^2$ (from $y = 0$ to $y = 4$) about the $y$-axis.

Solution: Since we revolve around the $y$-axis, we slice perpendicular to the $y$-axis, giving disks in $y$. Express the curve as $x = \sqrt{y}$, so each disk has radius $R(y) = \sqrt{y}$. $$V = \pi \int_0^4 \bigl(\sqrt{y}\bigr)^2\,dy = \pi \int_0^4 y\,dy = \pi\left[\frac{y^2}{2}\right]_0^4 = 8\pi$$
Common Pitfall: Students often forget to square the radius functions before integrating, or they square the entire integrand $R^2 - r^2$ as $(R - r)^2$ instead of computing $R^2 - r^2$ separately. Remember: the washer area is $\pi(R^2 - r^2)$, not $\pi(R - r)^2$. These give different results whenever $r \neq 0$.

Disk method: y = √x revolved about the x-axis. Use the slider to move the representative disk along the curve and observe how radius R(x) = √x changes.

Disk method: revolving y = √x about the x-axis. Each cross-section is a disk of radius R(x) = √x.

8.3 Volumes of Revolution: Shell Method BC

The shell method provides an alternative approach to computing volumes of revolution. Instead of slicing perpendicular to the axis of rotation (as with disks/washers), we slice parallel to the axis, producing thin cylindrical shells. Each shell is "unrolled" into a flat rectangular slab whose area is $2\pi \cdot \text{radius} \cdot \text{height}$.

Shell Method (Revolution About the $y$-axis)
When the region under $y = f(x)$ from $x = a$ to $x = b$ (where $0 \le a < b$) is revolved about the $y$-axis: $$V = 2\pi \int_a^b x \cdot f(x)\,dx$$ Here $x$ is the shell radius (distance from the shell to the $y$-axis) and $f(x)$ is the shell height.

The shell method is particularly useful when the disk/washer setup would require solving for $x$ in terms of $y$ and the resulting expression is complicated, or when the region would need to be split into multiple integrals using the washer approach but can be handled in a single integral using shells.

When to Choose Shell vs. Washer:
Example 8.3.1: Find the volume of the solid formed by revolving $y = x^2$, from $x = 0$ to $x = 2$, about the $y$-axis using the shell method.

Solution: Each cylindrical shell at position $x$ has radius $x$ and height $f(x) = x^2$. $$V = 2\pi \int_0^2 x \cdot x^2\,dx = 2\pi \int_0^2 x^3\,dx = 2\pi\left[\frac{x^4}{4}\right]_0^2 = 2\pi \cdot \frac{16}{4} = 8\pi$$ Note: this matches Example 8.2.3, which found the same volume using the disk method in $y$. Both methods yield $8\pi$.
Example 8.3.2: Find the volume of the solid formed by revolving the region between $y = x$ and $y = x^2$ about the $y$-axis.

Solution: The curves intersect at $x = 0$ and $x = 1$, with $x \ge x^2$ on $[0,1]$. Each shell at position $x$ has radius $x$ and height $(x - x^2)$. $$V = 2\pi \int_0^1 x\bigl(x - x^2\bigr)\,dx = 2\pi \int_0^1 \bigl(x^2 - x^3\bigr)\,dx$$ $$= 2\pi\left[\frac{x^3}{3} - \frac{x^4}{4}\right]_0^1 = 2\pi\left(\frac{1}{3} - \frac{1}{4}\right) = 2\pi \cdot \frac{1}{12} = \frac{\pi}{6}$$

8.4 Volumes with Known Cross Sections

Not all volume problems involve revolving a region. Many AP Calculus problems ask you to build a solid whose base is a known planar region and whose cross sections perpendicular to an axis have a specified geometric shape (squares, semicircles, equilateral triangles, etc.). The volume is found by integrating the cross-sectional area.

Volume by Cross Sections
If the cross section perpendicular to the $x$-axis at position $x$ has area $A(x)$, then: $$V = \int_a^b A(x)\,dx$$ The critical step is expressing $A(x)$ in terms of $x$ using the geometry of the cross-sectional shape and the dimensions of the base region.

Here is a quick reference for common cross-sectional shapes. Let $s(x)$ denote the side length (or diameter) of the cross section at position $x$, which is typically the vertical distance between the curves that form the base:

Example 8.4.1 (Square Cross Sections): The base of a solid is the region between $y = \sqrt{x}$ and $y = 0$ from $x = 0$ to $x = 4$. Cross sections perpendicular to the $x$-axis are squares. Find the volume.

Solution: At each $x$, the side of the square equals the distance from $y = 0$ to $y = \sqrt{x}$, so $s(x) = \sqrt{x}$. The area of each square cross section is: $$A(x) = [\sqrt{x}]^2 = x$$ $$V = \int_0^4 x\,dx = \left[\frac{x^2}{2}\right]_0^4 = \frac{16}{2} = 8$$
Example 8.4.2 (Semicircular Cross Sections): The base of a solid is the region between $y = x$ and $y = x^2$ from $x = 0$ to $x = 1$. Cross sections perpendicular to the $x$-axis are semicircles with diameter running from one curve to the other. Find the volume.

Solution: The diameter at position $x$ is $s(x) = x - x^2$. The area of a semicircle with diameter $s$ is $A = \frac{\pi}{8}s^2$: $$A(x) = \frac{\pi}{8}(x - x^2)^2 = \frac{\pi}{8}(x^2 - 2x^3 + x^4)$$ $$V = \frac{\pi}{8}\int_0^1 (x^2 - 2x^3 + x^4)\,dx = \frac{\pi}{8}\left[\frac{x^3}{3} - \frac{x^4}{2} + \frac{x^5}{5}\right]_0^1$$ $$= \frac{\pi}{8}\left(\frac{1}{3} - \frac{1}{2} + \frac{1}{5}\right) = \frac{\pi}{8}\cdot\frac{10 - 15 + 6}{30} = \frac{\pi}{8}\cdot\frac{1}{30} = \frac{\pi}{240}$$
AP Tip: Cross-section problems appear almost every year on the AP free-response section. The most common mistake is confusing the side length with the diameter. Read the problem carefully: if the cross section is a semicircle, determine whether the given length is the diameter or the radius. It is almost always the diameter.

Cross sections with known shape: circular base x²+y²=r² (adjust r with slider) with side length s(x) = 2√(r²−x²). The slider shows the base curve and the side length at position x₀.

Cross sections: circular base x²+y²=4 with square cross-sections perpendicular to the x-axis. Side s(x) = 2√(4−x²).

8.5 Arc Length BC

The arc length formula measures the distance along a curve. It extends the Pythagorean theorem to infinitesimal segments: each tiny piece of the curve has horizontal run $dx$ and vertical rise $dy = f'(x)\,dx$, so its length is $\sqrt{(dx)^2 + (dy)^2} = \sqrt{1 + [f'(x)]^2}\,dx$.

Arc Length Formula
If $f'(x)$ is continuous on $[a, b]$, the length of the curve $y = f(x)$ from $x = a$ to $x = b$ is: $$L = \int_a^b \sqrt{1 + \bigl[f'(x)\bigr]^2}\,dx$$

Arc length integrals are notoriously difficult to evaluate by hand. On the AP exam, you may be asked to set up the integral (without evaluating) or to evaluate it for a carefully chosen function where the algebra simplifies. Functions that produce clean arc length integrals often involve terms like $x^{3/2}$, $x^2$, or $e^x$ that pair nicely inside the square root.

Example 8.5.1: Find the arc length of $y = \frac{x^{3/2}}{3}$ from $x = 0$ to $x = 5$.

Solution: Compute the derivative: $$f'(x) = \frac{3}{2}\cdot\frac{x^{1/2}}{3} = \frac{\sqrt{x}}{2}$$ Then $[f'(x)]^2 = \frac{x}{4}$, so: $$L = \int_0^5 \sqrt{1 + \frac{x}{4}}\,dx = \int_0^5 \sqrt{\frac{4 + x}{4}}\,dx = \frac{1}{2}\int_0^5 \sqrt{4 + x}\,dx$$ Let $u = 4 + x$, $du = dx$. When $x = 0$, $u = 4$; when $x = 5$, $u = 9$. $$L = \frac{1}{2}\int_4^9 \sqrt{u}\,du = \frac{1}{2}\cdot\frac{2}{3}\bigl[u^{3/2}\bigr]_4^9 = \frac{1}{3}\bigl(27 - 8\bigr) = \frac{19}{3}$$

8.6 Average Value of a Function

The average value of a continuous function over an interval generalizes the familiar concept of an arithmetic mean to a function that takes infinitely many values. Intuitively, it is the height of the rectangle that has the same area as the region under the curve.

Average Value of a Function
If $f$ is continuous on $[a, b]$, the average value of $f$ on $[a, b]$ is: $$f_{\text{avg}} = \frac{1}{b - a}\int_a^b f(x)\,dx$$
Mean Value Theorem for Integrals
If $f$ is continuous on $[a, b]$, there exists at least one $c$ in $(a, b)$ such that: $$f(c) = \frac{1}{b - a}\int_a^b f(x)\,dx$$ In other words, the function actually attains its average value at some point in the interval.

This theorem guarantees that for any continuous function, there is always a point where the function equals its average. On the AP exam, you may be asked to find the value of $c$ where this occurs.

Example 8.6.1: Find the average value of $f(x) = x^2$ on the interval $[0, 3]$.

Solution: $$f_{\text{avg}} = \frac{1}{3 - 0}\int_0^3 x^2\,dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3}\cdot\frac{27}{3} = \frac{9}{3} = 3$$ By the Mean Value Theorem for Integrals, $f(c) = 3$ for some $c$ in $(0, 3)$. Setting $c^2 = 3$ gives $c = \sqrt{3} \approx 1.732$.
Example 8.6.2: Find the average value of $f(x) = \sin x$ on the interval $[0, \pi]$, and find $c$ where $f(c) = f_{\text{avg}}$.

Solution: $$f_{\text{avg}} = \frac{1}{\pi - 0}\int_0^{\pi}\sin x\,dx = \frac{1}{\pi}\bigl[-\cos x\bigr]_0^{\pi} = \frac{1}{\pi}\bigl[-\cos\pi + \cos 0\bigr] = \frac{1}{\pi}(1 + 1) = \frac{2}{\pi}$$ To find $c$: set $\sin c = \frac{2}{\pi}$, so $c = \arcsin\!\left(\frac{2}{\pi}\right) \approx 0.690$ radians. Since $\frac{2}{\pi} \approx 0.6366 < 1$, such a $c$ exists in $(0, \pi)$.
AP Tip: The average value formula is tested frequently on both multiple-choice and free-response sections. Do not confuse it with the average rate of change $\frac{f(b) - f(a)}{b - a}$. The average value uses an integral; the average rate of change uses a difference quotient.

8.7 Practice Problems

Test your understanding with these ten problems covering every topic in this chapter. Try each problem on your own before expanding the solution.

Problem 1. Find the area of the region enclosed by $y = 4 - x^2$ and $y = x^2 - 2x$.
Show Solution
Set $4 - x^2 = x^2 - 2x$: $4 = 2x^2 - 2x$, so $2x^2 - 2x - 4 = 0$, $x^2 - x - 2 = 0$, $(x-2)(x+1) = 0$. Intersections at $x = -1$ and $x = 2$.

On $[-1, 2]$, check $x = 0$: $4 - 0 = 4 > 0 - 0 = 0$, so $y = 4 - x^2$ is on top. $$A = \int_{-1}^{2}\bigl[(4 - x^2) - (x^2 - 2x)\bigr]\,dx = \int_{-1}^{2}(4 + 2x - 2x^2)\,dx$$ $$= \left[4x + x^2 - \frac{2x^3}{3}\right]_{-1}^{2} = \left(8 + 4 - \frac{16}{3}\right) - \left(-4 + 1 + \frac{2}{3}\right)$$ $$= \frac{20}{3} - \left(-\frac{7}{3}\right) = \frac{20}{3} + \frac{7}{3} = 9$$
Problem 2. Find the volume of the solid formed by revolving the region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 9$ about the $x$-axis.
Show Solution
Using the disk method with $R(x) = \sqrt{x}$: $$V = \pi\int_0^9 (\sqrt{x})^2\,dx = \pi\int_0^9 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^9 = \frac{81\pi}{2}$$
Problem 3. The region between $y = x$ and $y = x^2$ is revolved about the line $y = -1$. Find the volume using the washer method.
Show Solution
The curves intersect at $x = 0$ and $x = 1$. The axis of rotation is $y = -1$.

Outer radius: $R(x) = x - (-1) = x + 1$ (distance from $y = x$ to $y = -1$).
Inner radius: $r(x) = x^2 - (-1) = x^2 + 1$ (distance from $y = x^2$ to $y = -1$).

$$V = \pi\int_0^1\bigl[(x+1)^2 - (x^2+1)^2\bigr]\,dx$$ Expand: $(x+1)^2 = x^2 + 2x + 1$ and $(x^2+1)^2 = x^4 + 2x^2 + 1$. $$= \pi\int_0^1 (x^2 + 2x + 1 - x^4 - 2x^2 - 1)\,dx = \pi\int_0^1(-x^4 - x^2 + 2x)\,dx$$ $$= \pi\left[-\frac{x^5}{5} - \frac{x^3}{3} + x^2\right]_0^1 = \pi\left(-\frac{1}{5} - \frac{1}{3} + 1\right) = \pi\cdot\frac{-3 - 5 + 15}{15} = \frac{7\pi}{15}$$
Problem 4. BC Use the shell method to find the volume of the solid formed by revolving the region bounded by $y = x^3$, $y = 0$, and $x = 1$ about the $y$-axis.
Show Solution
Shell radius $= x$, shell height $= x^3$. $$V = 2\pi\int_0^1 x \cdot x^3\,dx = 2\pi\int_0^1 x^4\,dx = 2\pi\left[\frac{x^5}{5}\right]_0^1 = \frac{2\pi}{5}$$
Problem 5. The base of a solid is the region between $y = 1 - x^2$ and $y = 0$. Cross sections perpendicular to the $x$-axis are equilateral triangles. Find the volume.
Show Solution
The base region extends from $x = -1$ to $x = 1$. The side of each equilateral triangle is $s(x) = 1 - x^2$. The area of an equilateral triangle with side $s$ is $A = \frac{\sqrt{3}}{4}s^2$. $$V = \frac{\sqrt{3}}{4}\int_{-1}^{1}(1 - x^2)^2\,dx = \frac{\sqrt{3}}{4}\int_{-1}^{1}(1 - 2x^2 + x^4)\,dx$$ Since the integrand is even, we can write this as: $$= \frac{\sqrt{3}}{4}\cdot 2\int_0^1(1 - 2x^2 + x^4)\,dx = \frac{\sqrt{3}}{2}\left[x - \frac{2x^3}{3} + \frac{x^5}{5}\right]_0^1$$ $$= \frac{\sqrt{3}}{2}\left(1 - \frac{2}{3} + \frac{1}{5}\right) = \frac{\sqrt{3}}{2}\cdot\frac{15 - 10 + 3}{15} = \frac{\sqrt{3}}{2}\cdot\frac{8}{15} = \frac{4\sqrt{3}}{15}$$
Problem 6. Find the average value of $f(x) = 3x^2 + 1$ on $[1, 4]$.
Show Solution
$$f_{\text{avg}} = \frac{1}{4 - 1}\int_1^4(3x^2 + 1)\,dx = \frac{1}{3}\left[x^3 + x\right]_1^4 = \frac{1}{3}\bigl[(64 + 4) - (1 + 1)\bigr] = \frac{1}{3}(66) = 22$$
Problem 7. BC Set up (but do not evaluate) the arc length integral for $y = \ln(\cos x)$ from $x = 0$ to $x = \frac{\pi}{4}$.
Show Solution
Compute $f'(x) = \frac{-\sin x}{\cos x} = -\tan x$. Then $[f'(x)]^2 = \tan^2 x$. $$L = \int_0^{\pi/4}\sqrt{1 + \tan^2 x}\,dx = \int_0^{\pi/4}\sqrt{\sec^2 x}\,dx = \int_0^{\pi/4}\sec x\,dx$$ (This integral evaluates to $\ln|\sec x + \tan x|\Big|_0^{\pi/4} = \ln(\sqrt{2} + 1)$, but the problem asks only for setup.)
Problem 8. Find the volume of the solid formed by revolving the region bounded by $y = e^x$, $y = 1$, and $x = \ln 3$ about the $x$-axis.
Show Solution
The region lies between $y = e^x$ (outer) and $y = 1$ (inner) from $x = 0$ (where $e^x = 1$) to $x = \ln 3$. Using washers: $$V = \pi\int_0^{\ln 3}\bigl[(e^x)^2 - 1^2\bigr]\,dx = \pi\int_0^{\ln 3}(e^{2x} - 1)\,dx$$ $$= \pi\left[\frac{e^{2x}}{2} - x\right]_0^{\ln 3} = \pi\left[\left(\frac{e^{2\ln 3}}{2} - \ln 3\right) - \left(\frac{1}{2} - 0\right)\right]$$ Since $e^{2\ln 3} = (e^{\ln 3})^2 = 9$: $$= \pi\left(\frac{9}{2} - \ln 3 - \frac{1}{2}\right) = \pi(4 - \ln 3)$$
Problem 9. The base of a solid is the region enclosed by $y = \sin x$ and $y = 0$ from $x = 0$ to $x = \pi$. Cross sections perpendicular to the $x$-axis are squares. Find the volume.
Show Solution
Each square has side $s(x) = \sin x$, so $A(x) = \sin^2 x$. $$V = \int_0^{\pi}\sin^2 x\,dx = \int_0^{\pi}\frac{1 - \cos 2x}{2}\,dx = \frac{1}{2}\left[x - \frac{\sin 2x}{2}\right]_0^{\pi}$$ $$= \frac{1}{2}\left(\pi - 0\right) = \frac{\pi}{2}$$
Problem 10. BC Use the shell method to find the volume of the solid formed by revolving the region between $y = 4x - x^2$ and $y = 0$ about the line $x = 5$.
Show Solution
The parabola $y = 4x - x^2 = x(4 - x)$ has roots at $x = 0$ and $x = 4$. Revolving about $x = 5$, each shell at position $x$ has:
Radius $= 5 - x$ (distance from $x$ to $x = 5$)
Height $= 4x - x^2$ $$V = 2\pi\int_0^4 (5 - x)(4x - x^2)\,dx = 2\pi\int_0^4 (20x - 5x^2 - 4x^2 + x^3)\,dx$$ $$= 2\pi\int_0^4(20x - 9x^2 + x^3)\,dx = 2\pi\left[10x^2 - 3x^3 + \frac{x^4}{4}\right]_0^4$$ $$= 2\pi\left(160 - 192 + 64\right) = 2\pi(32) = 64\pi$$
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