International Baccalaureate · DP 2: Applications and Interpretation HL

Areas and Volumes

⌚ ~59 min ▢ 7 blocks ✓ 30 exercises ➣ Prerequisites : Integration

I Areas

I.1 The Area of a Region

The definite integral $\int_a^b f(x)\,\mathrm{d}x$ is the signed area between the graph of $y=f(x)$ and the $x$-axis. This means that areas above the x-axis are positive, while areas below are negative.
However, when a problem asks for the geometric area or simply the area of a region, it refers to the physical, positive space the region occupies. In this case, we must ensure that all parts of the region, whether they are above or below the x-axis, contribute a positive value to the total. This total geometric area needs its own method.

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Method — Calculating Geometric Area Bounded by a Curve and the x-axis

To find the total geometric area $\mathcal{A}$ bounded by a curve $y=f(x)$ and the x-axis from $x=a$ to $x=b$ ($f$ continuous on $[a, b]$, with finitely many roots in this interval):

Find the $x$-intercepts: Determine where the curve meets the $x$-axis (the zeros of $f$) by solving $f(x)=0$. Identify any roots $c_1, c_2, \dots$ that lie within the interval $[a,b]$.
Split the integral: Divide the main integral into smaller integrals at each intercept found in the previous step: on each sub-interval, $f$ keeps a constant sign (it is continuous and does not vanish inside).
Calculate each definite integral: Compute the integral for each sub-interval. Some of these will correspond to positive values (where $f(x) \ge 0$) and some to negative values (where $f(x) \le 0$).
Sum the absolute values: The total geometric area is the sum of the absolute values of these integrals. If the integral of a sub-region is negative, take its positive value before adding it to the total.

Example

Find the total geometric area between the curve $y=x^2-1$ and the $x$-axis from $x=0$ to $x=2$.

Answer

Find intercepts: We solve $f(x)=x^2-1=0$, which gives roots at $x=1$ and $x=-1$. The only root within our interval of integration $[0,2]$ is $x=1$.
Split the integral: We must split the total area calculation at $x=1$.
- From $x=0$ to $x=1$, the function is below the $x$-axis (Area $\mathcal{A}_1$).
- From $x=1$ to $x=2$, the function is above the $x$-axis (Area $\mathcal{A}_2$).
Calculate each integral: $$ \int_0^1 (x^2-1)\, \mathrm{d}x = \left[\frac{x^3}{3}-x\right]_0^1 = \left(\frac{1}{3}-1\right) - 0 = -\frac{2}{3} $$ $$ \int_1^2 (x^2-1)\, \mathrm{d}x = \left[\frac{x^3}{3}-x\right]_1^2 = \left(\frac{8}{3}-2\right) - \left(\frac{1}{3}-1\right) = \frac{2}{3} - \left(-\frac{2}{3}\right) = \frac{4}{3} $$
Sum the absolute values: $$ \text{Total Area} = \left|-\frac{2}{3}\right| + \left|\frac{4}{3}\right| = \frac{2}{3} + \frac{4}{3} = \frac{6}{3} = 2 $$

The total geometric area is 2 square units.

Definition — Geometric Area Formula

The total geometric area $\mathcal{A}$ between the graph of a function $f(x)$ and the $x$-axis from $x=a$ to $x=b$ is given by the integral of the absolute value of the function: $$ \mathcal{A} = \int_a^b |f(x)|\, \mathrm{d}x $$

Note

The method of splitting the integral and summing the absolute values is equivalent to this formal definition of the total geometric area.

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Skills to practice

Evaluating Areas with Geometric Formulas
Finding Areas with Antiderivatives

I.2 The Area Between a Curve and the $y$-Axis

When a region is bounded by a curve and the $y$-axis, it is natural to integrate with respect to $y$: express $x$ as a function of $y$, $x = g(y)$, and compute $\int g(y)\,\mathrm{d}y$. For a region to the right of the $y$-axis, $x = g(y) \ge 0$, and this integral is the area; for a region to the left of the axis, $g(y) \le 0$, and the area is $\int |g(y)|\,\mathrm{d}y$.

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Method — Area Between a Curve and the y-Axis

Express $x$ as a function of $y$: solve the equation of the curve for $x=g(y)$.
Identify the $y$-bounds $c$ and $d$ of the region.
Integrate with respect to $y$: $\displaystyle\mathcal{A}=\int_c^d g(y)\,\mathrm{d}y$ (for $g(y)\ge0$).

Example

Find the area of the region bounded by the curve $y=x^2$ (with $x\ge0$), the $y$-axis, and the line $y=4$.

Answer

For $x\ge0$, $y=x^2$ gives $x=\sqrt{y}$. The shaded region is bounded by the curve, the $y$-axis, and the line $y=4$:

Integrating with respect to $y$ from $0$ to $4$: $$\begin{aligned} \mathcal{A} &= \int_0^4 \sqrt{y}\,\mathrm{d}y &&\text{(area vs the }y\text{-axis)}\\ &= \Big[\tfrac23 y^{3/2}\Big]_0^4 = \tfrac{16}{3} &&\text{(evaluate)} \end{aligned}$$

Skills to practice

Calculating Areas Against the $y$-Axis

II Volumes of Revolution

II.1 Volumes of Revolution about the $x$-Axis

A solid of revolution is a three-dimensional object formed by rotating a two-dimensional shape around an axis. Their exact volume can be found by integration.
Consider the area under the curve $y=f(x)$ from $x=a$ to $x=b$. If we revolve this area $360^\circ$ ($2\pi$ radians) around the $x$-axis, it sweeps out a solid of revolution.

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Method — The Disc Method

To find the volume of this solid, we use the same strategy as for area: slice the solid into many thin pieces and sum their volumes. In this case, each slice is a thin cylindrical disc. The key insight is that each disc is formed by rotating one of the thin rectangles from a Riemann sum around the $x$-axis.

The volume of a single cylinder is $\pi r^2 h$. For a disc at position $x_i$ with a small thickness $\Delta x$:

The radius is the height of the function, $r = f(x_i)$.
The height (thickness) of the disc is $h = \Delta x$.

The volume of one disc is $V_i = \pi [f(x_i)]^2 \Delta x$. The total volume is approximated by summing the volumes of all the discs: $$\begin{aligned} \mathcal{V}&\approx \sum_{i=0}^{n-1}V_i\\ &\approx \sum_{i=0}^{n-1} \pi [f(x_i)]^2 \Delta x\\ \end{aligned} $$ To find the volume exactly, we take the limit as the number of discs goes to infinity and their thickness approaches zero ($\Delta x \to 0$). This limit is the definite integral: $$\begin{aligned} \mathcal{V}&= \lim_{n \to \infty} \sum_{i=0}^{n-1} \pi [f(x_i)]^2 \Delta x\\ &= \int_a^b \pi [f(x)]^2\, \mathrm{d}x\\ &=\pi \int_a^b [f(x)]^2\, \mathrm{d}x\\ \end{aligned} $$

Proposition — Volume of Revolution about the $x$-axis

Let $f$ be continuous on $[a,b]$, with $f(x)\geq 0$ on $[a,b]$. The volume $V$ generated by rotating the region bounded by the curve $y=f(x)$, the $x$-axis, and the lines $x=a$ and $x=b$ around the $x$-axis is given by: $$ V = \pi \int_a^b [f(x)]^2\, \mathrm{d}x $$

Example

Find the volume of the solid generated by revolving the region under the curve $y=\sqrt{x}$ from $x=1$ to $x=4$ around the $x$-axis.

Answer

Identify: The function is $f(x)=\sqrt{x}$, and the limits are $a=1$ and $b=4$.
Square the function: $[f(x)]^2 = (\sqrt{x})^2 = x$.
Integrate: We set up and evaluate the integral for the volume: $$\begin{aligned} V &= \pi \int_1^4 x \, \mathrm{d}x \\ &= \pi \left[ \frac{x^2}{2} \right]_1^4 \\ &= \pi \left( \frac{4^2}{2} - \frac{1^2}{2} \right) \\ &= \pi \left( \frac{16}{2} - \frac{1}{2} \right) \\ &= \pi \left( 8 - \frac{1}{2} \right)\\ &= \frac{15\pi}{2} \end{aligned}$$

The volume of the solid is $\frac{15\pi}{2}$ cubic units.

Skills to practice

Calculating Volumes of Revolution about the $x$-Axis

II.2 Volumes of Revolution about the $y$-Axis

Many objects have a vertical axis of symmetry: a vase, a bowl, a funnel. Their volume is a volume of revolution about the $y$-axis. The slices are then horizontal discs, of thickness $\mathrm{d}y$, and the radius of the disc at height $y$ is the horizontal distance $x$ from the axis to the curve.

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Proposition — Volume of Revolution about the y-axis

Let $g$ be continuous on $[c,d]$, with $g(y)\ge 0$ on $[c,d]$. The volume $V$ generated by rotating the region bounded by the curve $x=g(y)$, the y-axis, and the lines $y=c$ and $y=d$ around the y-axis is given by: $$ V = \pi \int_c^d [g(y)]^2\, \mathrm{d}y $$

Note

To use this formula, the function must be expressed in the form $x=g(y)$, where $y$ is the independent variable. This may require rearranging the function's equation.

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Example

Find the volume of the solid generated by revolving the region bounded by the curve $y=\sqrt{x}$, the y-axis, and the line $y=2$ about the y-axis.

Answer

Rearrange the function: The rotation is around the y-axis, so we need to express $x$ in terms of $y$: $$ y = \sqrt{x} \implies x = y^2. $$ So, our function is $g(y)=y^2$.
Identify limits: The region is bounded by the y-axis (which corresponds to $x=0$ and starts at $y=0$, where the curve meets the axis) and the line $y=2$. So, our limits are $c=0$ and $d=2$.
Integrate: We set up and evaluate the integral for the volume: $$\begin{aligned} V &= \pi \int_0^2 [g(y)]^2\, \mathrm{d}y \\ &= \pi \int_0^2 (y^2)^2\, \mathrm{d}y \\ &= \pi \int_0^2 y^4\, \mathrm{d}y \\ &= \pi \left[ \frac{y^5}{5} \right]_0^2 \\ &= \pi \left( \frac{2^5}{5} - \frac{0^5}{5} \right)\\ &= \frac{32\pi}{5} \end{aligned}$$

The volume of the solid is $\frac{32\pi}{5}$ cubic units.

Skills to practice

Calculating Volumes of Revolution about the $y$-Axis

II.3 Volumes of Revolution with Technology

Method — Finding a Volume of Revolution with Technology

A graphing calculator evaluates the integral of a volume of revolution once the volume is written.

Write the volume first. For the region between a curve and the axis of rotation: about the $x$-axis, $V = \pi\displaystyle\int_a^b [f(x)]^2\,\mathrm{d}x$; about the $y$-axis, $V = \pi\displaystyle\int_c^d [g(y)]^2\,\mathrm{d}y$, the curve being written $x = g(y)$. The function is squared, not the integral, and the factor $\pi$ multiplies the whole integral.
Simplify the square when it is quick, for example $\left(\sqrt{x}\right)^2 = x$ or $\left(e^{-x^2}\right)^2 = e^{-2x^2}$; otherwise enter it as it is, with brackets around the function.
Enter the integral with the variable $X$ of the calculator, $\pi$ typed in front of it. For a rotation about the $y$-axis, the variable $y$ is typed as $X$: the name of the variable of integration does not change the value of the integral.
Report the volume to 3 significant figures.

The mock-up evaluates the volume generated by the region under $y = e^{-x^2}$, for $0 \le x \le 1.5$, rotated about the $x$-axis.

Example

The region bounded by the curve $y = e^{-x^2}$, the $x$-axis, the $y$-axis and the line $x = 1.5$ is rotated about the $x$-axis. Find the volume of the solid generated.

Answer

For $0 \le x \le 1.5$, $e^{-x^2} > 0$, so the volume is written first, then the square simplified: $$\begin{aligned}[t] V &= \pi\int_0^{1.5} \left(e^{-x^2}\right)^2\mathrm{d}x &&\text{(volume about the } x\text{-axis)}\\ &= \pi\int_0^{1.5} e^{-2x^2}\,\mathrm{d}x. &&\big((e^{u})^2 = e^{2u}\big) \end{aligned}$$ No antiderivative of $e^{-2x^2}$ can be written with the standard functions: the calculator evaluates the integral. The template, with $\pi$ in front, the integrand $e^{-2X^2}$ and the bounds $0$ and $1.5$, displays $1.9633862$. The volume is $V \approx 1.96$ cubic units.
Two entries give a wrong value: $\pi\displaystyle\int_0^{1.5} e^{-x^2}\,\mathrm{d}x$, where the function is not squared, displays $2.6897952$; the integral without the factor $\pi$ displays $0.6249652$.

Skills to practice

Finding Volumes of Revolution with Technology

III Going further

Skills to practice

Solving Problems