Area And Volume From Double Integrals

From a Planar Region to Area and Volume

A double integral can encode a geometric quantity, not just the accumulation of a function’s values. When the function is the constant one, the integral measures the area of the region. When the function gives a nonnegative height above the region, the integral measures the volume beneath its graph.

The essential point is that the region and the integrand play different roles. The region determines where the accumulation takes place; the integrand determines what is accumulated at each point. For area, each point contributes one unit of area per unit of planar area. For volume, each point contributes its height times a small piece of planar area.

Definition: For a bounded planar region $R$ whose boundary is sufficiently regular for its area to be defined by integration, its area is $$ A(R)=\iint_R 1\,dA. $$ If $h:R\to\mathbb{R}$ is a continuous, nonnegative height function, the volume of the solid between $z=0$ and $z=h(x,y)$, above $R$, is $$ V=\iint_R h(x,y)\,dA. $$

Here $dA$ denotes a small element of planar area. In Cartesian coordinates, $dA=dy\,dx$ or $dA=dx\,dy$, according to the order of integration. In polar coordinates, $dA=r\,dr\,d\theta$. The factor $r$ in the polar area element is essential.

Why the Integrals Give Area and Volume

Imagine dividing $R$ into many small pieces, each with area $\Delta A_i$, and selecting a point $(x_i,y_i)$ in each piece. The sum $\sum_i \Delta A_i$ approximates the area of $R$. It is also a sum of the constant value one times each piece’s area:

\sum_i 1\cdot\Delta A_i.

As the pieces become smaller, this sum approaches $\iint_R 1\,dA$. For volume, the piece above the $i$-th planar piece is approximately a column of height $h(x_i,y_i)$ and base area $\Delta A_i$. Its volume is approximately $h(x_i,y_i)\Delta A_i$. Summing the column volumes and refining the division gives the double integral of $h$.

\sum_i h(x_i,y_i)\Delta A_i \longrightarrow \iint_R h(x,y)\,dA.

This also explains the units. If coordinates are measured in metres, then $dA$ has units of square metres, so $\iint_R 1\,dA$ is an area. If $h$ is measured in metres, then $h\,dA$ has units of cubic metres, so its integral is a volume.

Writing the Bounds as Iterated Integrals

A region may be described by letting one coordinate vary over an interval and giving the other coordinate’s lower and upper limits. For example, if $R=\{(x,y):a\le x\le b,\ g_1(x)\le y\le g_2(x)\}$, then integration with respect to $y$ first gives

\iint_R f(x,y)\,dA = \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx.

For area, set $f(x,y)=1$; for volume beneath a height function, set $f(x,y)=h(x,y)$. Alternatively, if the region is described by $c\le y\le d$ and $q_1(y)\le x\le q_2(y)$, use

\iint_R f(x,y)\,dA = \int_c^d\int_{q_1(y)}^{q_2(y)} f(x,y)\,dx\,dy.

These iterated forms apply when the region and integrand meet the usual conditions for changing a double integral into an iterated integral—for example, a bounded region with piecewise-smooth boundary and a continuous integrand. The limits must describe the region exactly, with no gaps or unintended extra portions.

Worked Examples

Worked Example: Area Between Two Curves

Find the area enclosed between $y=x^2$ and $y=x$ for $0\le x\le1$. First, the curves meet where $x^2=x$, so $x(x-1)=0$ and the intersection values are $x=0$ and $x=1$. For $0\le x\le1$, $x^2\le x$, so the lower limit is $x^2$ and the upper limit is $x$.

\begin{aligned} A &=\int_0^1\int_{x^2}^{x}1\,dy\,dx\\ &=\int_0^1 [y]_{x^2}^{x}\,dx\\ &=\int_0^1(x-x^2)\,dx\\ &=\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1\\ &=\left(\frac12-\frac13\right)-0\\ &=\frac16. \end{aligned}

To check the result, reverse the order of integration. The region has $0\le y\le1$; at fixed $y$, its $x$-values run from $x=y$ to $x=\sqrt y$. Thus

\begin{aligned} A &=\int_0^1\int_y^{\sqrt y}1\,dx\,dy\\ &=\int_0^1(\sqrt y-y)\,dy\\ &=\left[\frac23y^{3/2}-\frac12y^2\right]_0^1\\ &=\frac23-\frac12\\ &=\frac16. \end{aligned}

Both orders give the same area. The nonnegative width $x-x^2$ on $0\le x\le1$ is also consistent with the region’s bounds.

Worked Example: Volume Above a Triangular Region

Let $R$ be the triangle $0\le x\le2,\ 0\le y\le2-x$, and let the height be $h(x,y)=x+y$. Since $x\ge0$ and $y\ge0$ on $R$, the height is nonnegative. The volume is

\begin{aligned} V &=\int_0^2\int_0^{2-x}(x+y)\,dy\,dx\\ &=\int_0^2\left[xy+\frac{y^2}{2}\right]_0^{2-x}\,dx\\ &=\int_0^2\left(x(2-x)+\frac{(2-x)^2}{2}\right)\,dx\\ &=\int_0^2\left(2x-x^2+2-2x+\frac{x^2}{2}\right)\,dx\\ &=\int_0^2\left(2-\frac{x^2}{2}\right)\,dx\\ &=\left[2x-\frac{x^3}{6}\right]_0^2\\ &=4-\frac{8}{6}\\ &=\frac83. \end{aligned}

For a check, describe the same triangle by $0\le y\le2,\ 0\le x\le2-y$. Then

\begin{aligned} V &=\int_0^2\int_0^{2-y}(x+y)\,dx\,dy\\ &=\int_0^2\left[\frac{x^2}{2}+yx\right]_0^{2-y}\,dy\\ &=\int_0^2\left(\frac{(2-y)^2}{2}+y(2-y)\right)\,dy\\ &=\int_0^2\left(2-2y+\frac{y^2}{2}+2y-y^2\right)\,dy\\ &=\int_0^2\left(2-\frac{y^2}{2}\right)\,dy\\ &=\left[2y-\frac{y^3}{6}\right]_0^2\\ &=\frac83. \end{aligned}

The reversed bounds produce the same volume. In particular, the upper limit $2-x$ in the first integral and $2-y$ in the second both describe the sloping side $x+y=2$, rather than the entire square $0\le x,y\le2$.

Area and Volume with Polar Bounds

For a region described by $ \alpha\le\theta\le\beta$ and $r_{\min}(\theta)\le r\le r_{\max}(\theta)$, the polar area element is $r\,dr\,d\theta$. Consequently, area and volume take the forms

A(R)=\int_\alpha^\beta\int_{r_{\min}(\theta)}^{r_{\max}(\theta)}r\,dr\,d\theta, \qquad V=\int_\alpha^\beta\int_{r_{\min}(\theta)}^{r_{\max}(\theta)} h(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta.

The height function must be expressed at the point $(x,y)=(r\cos\theta,r\sin\theta)$. The additional factor $r$ comes from planar area, not from the height.

Worked Example: Area and Volume over an Annular Sector

Let $R$ be the first-quadrant annular sector $1\le r\le2,\ 0\le\theta\le\pi/2$. Find its area and the volume beneath $h(x,y)=\sqrt{x^2+y^2}$. On this region, $h=r$, so the polar integrals are

\begin{aligned} A(R) &=\int_0^{\pi/2}\int_1^2 r\,dr\,d\theta\\ &=\int_0^{\pi/2}\left[\frac{r^2}{2}\right]_1^2\,d\theta\\ &=\int_0^{\pi/2}\frac32\,d\theta\\ &=\frac{3\pi}{4}, \\[6pt] V &=\int_0^{\pi/2}\int_1^2 r\cdot r\,dr\,d\theta\\ &=\int_0^{\pi/2}\left[\frac{r^3}{3}\right]_1^2\,d\theta\\ &=\int_0^{\pi/2}\frac73\,d\theta\\ &=\frac{7\pi}{6}. \end{aligned}

For the area, the radial bounds give $(2^2-1^2)/2=3/2$, and multiplying by the angular width $\pi/2$ gives $3\pi/4$. For the volume, the height is $r$ and the area element contributes another $r$, so the radial integrand is $r^2$; its integral from 1 to 2 is $(8-1)/3=7/3$. This confirms that the height and area element have been included separately.

Conditions and a Common Mistake

To interpret $\iint_R h\,dA$ as the volume between the graph and the $xy$-plane, the height must satisfy $h(x,y)\ge0$ throughout $R$. If the function takes negative values, the integral computes signed volume: contributions below the $xy$-plane subtract from those above it. It is not automatically the total geometric volume of all parts of the solid.

Common mistake: In polar coordinates, writing $\iint_R 1\,dr\,d\theta$ for area omits the polar area factor. The correct area element is $r\,dr\,d\theta$, so omitting $r$ changes the quantity being integrated and generally gives the wrong result.

For instance, on the annular sector above, omitting $r$ would give $\int_0^{\pi/2}\int_1^2 1\,dr\,d\theta=\pi/2$, not the correct area $3\pi/4$. The discrepancy is not a rounding error: the incorrect integrand assigns equal planar area to equal changes in $r$ at different distances from the origin. The factor $r$ accounts for the widening of the region as the radius increases.

Choosing and Checking a Setup

Before evaluating, check that the integral matches the geometric quantity. Use $1$ for area and the specified nonnegative height for volume; in polar coordinates, include $r$ in addition. Then check that the limits describe the intended region and that the resulting units are appropriate. If the region can be described in the other integration order, evaluating that form is a useful independent check on both the limits and the calculation.

Check Your Understanding

Use the bounds and definitions in each question to set up an integral before evaluating it.

Write the area integral for the region $0\le x\le1,\ 0\le y\le x$.
For the same region and height $h(x,y)=2-y$, write the volume integral. Is the height nonnegative throughout the region?
Describe the region $0\le x\le1,\ x^2\le y\le x$ using horizontal slices, and write its area integral in the reversed order.
For $1\le r\le3,\ 0\le\theta\le\pi$, write the polar integral for the area. Which factor in the integrand accounts for the polar area element?
If a height function is negative on part of a region, what does its double integral represent, and why might that differ from the solid’s total geometric volume?