Tutorials › Calculus III › Deriving the Polar Area Element

Home  /  Tutorials  /  Polar Double Integrals

Polar Double Integrals · Tutorial 552 of 999

Deriving the Polar Area Element

Learn how the polar coordinate map stretches small areas and why a double integral uses the element \(r\,dr\,d\theta\).

Intermediate 9 min read

What You'll Learn

  • Write the polar coordinate map and compute its Jacobian determinant.
  • Explain how a derivative maps a small parameter rectangle to an approximate parallelogram.
  • Distinguish the local area element from the formula for the area of a whole region.
  • Identify when the polar map is one-to-one and where its Jacobian vanishes.
  • Check polar area calculations against exact geometric area.

Why Polar Coordinates Change Area

Polar coordinates describe a point in the plane by its distance \(r\) from the origin and its angle \(\theta\). A small change in either parameter does not generally produce the same-sized change in the plane: changing the angle sweeps out a longer arc when \(r\) is larger. Consequently, a small rectangle in the \(r,\theta\) parameter plane does not have the same area as its image in the \(x,y\) plane.

The factor that accounts for this change is \(r\). We will derive it from the polar coordinate map, then distinguish carefully between the local area element and the area of an entire region. The latter is an integral of the local factor; it is not, in general, obtained by multiplying parameter area by one fixed value of \(r\).

The Polar Coordinate Map and Its Jacobian

For \(r\ge 0\), the polar coordinate map \(T\) sends the parameter pair \((r,\theta)\) to its Cartesian coordinates:

$$ T(r,\theta)=(x,y)=(r\cos\theta,r\sin\theta). $$

The Jacobian matrix is the derivative of this map. Its columns record how the image point changes when \(r\) and \(\theta\) change, respectively:

$$ DT(r,\theta) = \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial\theta}\\[4pt] \frac{\partial y}{\partial r} & \frac{\partial y}{\partial\theta} \end{pmatrix} = \begin{pmatrix} \cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{pmatrix}. $$

Taking its determinant gives

$$ \det DT(r,\theta) = r\cos^2\theta+r\sin^2\theta = r. $$

Since \(r\ge 0\), the absolute value of the determinant is also \(r\). The determinant measures the local area scaling of the map. Its value depends on the radial coordinate: angular changes stretch more at larger radii.

Definition: The polar area element is the local area measure obtained by applying the absolute Jacobian determinant of the polar coordinate map to parameter area: $$ dA=|\det DT(r,\theta)|\,dr\,d\theta=r\,dr\,d\theta,\qquad r\ge 0. $$

Why the Determinant Gives an Area Factor

The derivative gives a linear approximation to a differentiable map near a point. A linear map sends a small rectangle to a parallelogram, and the absolute value of its determinant is the ratio of the parallelogram's area to the rectangle's area. Applying this fact to \(DT(r,\theta)\) explains why the local area scale is \(r\).

More explicitly, take a small parameter rectangle with side lengths \(\Delta r\) and \(\Delta\theta\), based near \((r,\theta)\). Its parameter area is \(\Delta r\,\Delta\theta\). Under the derivative, its two side directions are approximately the vectors \((\cos\theta,\sin\theta)\Delta r\) and \((-r\sin\theta,r\cos\theta)\Delta\theta\). The parallelogram area is the absolute determinant of these vectors:

$$ \left| \det \begin{pmatrix} \cos\theta\,\Delta r & -r\sin\theta\,\Delta\theta\\ \sin\theta\,\Delta r & r\cos\theta\,\Delta\theta \end{pmatrix} \right| = r\,\Delta r\,\Delta\theta. $$

For the actual curved image of the rectangle, this is a local approximation. As the rectangle shrinks, the differentiable map is increasingly well approximated by its derivative, so the ratio of image area to parameter area approaches \(r\). This local statement is what \(dA=r\,dr\,d\theta\) expresses.

Theorem: Let \(E\) be a parameter region and let \(T\) be a continuously differentiable, one-to-one change of variables on its interior, with a well-defined Jacobian determinant there. The area of its image is $$ \operatorname{area}(T(E)) = \iint_E |\det DT(u,v)|\,du\,dv. $$ For the polar map on a region with \(r\ge 0\), where it is one-to-one in the interior, this becomes $$ \operatorname{area}(T(E)) = \iint_E r\,dr\,d\theta. $$

The whole-region formula integrates the local factor over the parameter region. It does not say that the image area is the parameter area multiplied by one determinant value. That shortcut is valid only if the determinant is constant throughout the region.

Worked Example: Checking a Small Polar Patch

Worked Example: Checking a Small Polar Patch

Consider a parameter rectangle with \(2\le r\le 2.01\) and \(\pi/3\le\theta\le\pi/3+0.02\). Its radial width is \(\Delta r=0.01\), and its angular width is \(\Delta\theta=0.02\). Use the Jacobian to estimate the image area near \(r=2\), then compare it with the exact area of the curved patch.

At the inner radius \(r=2\), the local Jacobian estimate is

$$ r\,\Delta r\,\Delta\theta = 2(0.01)(0.02) = 0.0004. $$

The exact patch is an annular sector. A sector with angle \(\Delta\theta\) and radius \(R\) has area \(\frac12R^2\Delta\theta\). Subtract the inner sector from the outer one:

$$ \begin{aligned} \operatorname{area} &=\frac12\left((2.01)^2-2^2\right)(0.02)\\ &=\frac12(4.0401-4)(0.02)\\ &=\frac12(0.0401)(0.02)\\ &=0.000401. \end{aligned} $$

The exact area \(0.000401\) is close to the local estimate \(0.0004\), but not identical. The difference occurs because \(r\) increases across the patch, so the Jacobian is slightly larger than \(2\) except along its inner edge. As the patch becomes smaller, the local estimate becomes more accurate.

Check: The exact area is positive, as it must be, and is slightly larger than the estimate using only the inner radius. This agrees with the fact that the angular side of the patch is longer at larger \(r\).

From the Local Element to Total Area

Suppose a region is described in polar parameters by \(a\le r\le b\) and \(\alpha\le\theta\le\beta\), where \(0\le a<b\) and the angular interval gives a one-to-one description in the interior. The parameter region is a rectangle, but its image generally has curved boundaries. The change-of-variables formula gives its area by integrating the varying Jacobian:

$$ \operatorname{area} = \int_{\alpha}^{\beta}\int_a^b r\,dr\,d\theta. $$

The inner integration sums the local area contributions across the radial direction; the outer integration then sums across angles. The factor \(r\) must remain inside the integral because it varies with the radial position.

Worked Example: A Patch with a Varying Jacobian

Find the area of the image of \(1\le r\le 2\), \(0\le\theta\le\pi/3\). First integrate with respect to \(r\):

$$ \int_0^{\pi/3}\int_1^2 r\,dr\,d\theta = \int_0^{\pi/3} \left[\frac{r^2}{2}\right]_{1}^{2} \,d\theta = \int_0^{\pi/3}\frac{4-1}{2}\,d\theta. $$

The remaining integral is

$$ \int_0^{\pi/3}\frac32\,d\theta = \frac32\left[\theta\right]_0^{\pi/3} = \frac{\pi}{2}. $$

As a geometric check, the image is the difference between two sectors with the same angle \(\pi/3\), one of radius \(2\) and one of radius \(1\). Their area difference is

$$ \frac12(2^2)\frac{\pi}{3} - \frac12(1^2)\frac{\pi}{3} = \frac{2\pi}{3}-\frac{\pi}{6} = \frac{\pi}{2}. $$

The parameter rectangle has area \((2-1)(\pi/3)=\pi/3\), while its image has area \(\pi/2\). The determinant \(r\) varies from \(1\) to \(2\), so multiplying the parameter area by one value of \(r\) would not give the exact image area.

Worked Example: Verifying the Polar Area of a Quarter-Disk

Worked Example: Verifying the Polar Area of a Quarter-Disk

Consider \(0\le r\le 3\) and \(0\le\theta\le\pi/2\). The image is a quarter-disk of radius \(3\). Compute its area using the polar area element and compare it with one quarter of the circle's area.

Using the Jacobian factor gives

$$ \begin{aligned} \operatorname{area} &=\int_0^{\pi/2}\int_0^3 r\,dr\,d\theta\\ &=\int_0^{\pi/2}\left[\frac{r^2}{2}\right]_0^3\,d\theta\\ &=\int_0^{\pi/2}\frac92\,d\theta\\ &=\frac92\left[\theta\right]_0^{\pi/2}\\ &=\frac92\cdot\frac{\pi}{2}\\ &=\frac{9\pi}{4}. \end{aligned} $$

The full circle has area \(\pi(3)^2=9\pi\), so one quarter of it has area \(\frac14(9\pi)=\frac{9\pi}{4}\). The two calculations agree.

Check: The result is positive and one quarter of the full circle's area. If the Jacobian factor were omitted, the parameter rectangle would have area \(3\pi/2\), which is not the quarter-disk area.

Where the Formula Needs Care

The polar map has two features that matter when applying the change-of-variables formula. First, at \(r=0\), the Jacobian determinant is zero. All angles at \(r=0\) map to the origin, so the map is not one-to-one there. This single point has area zero; it can be excluded from the interior of a region without changing its area.

Second, angles differing by \(2\pi\) describe the same direction. A parameter region that includes multiple full turns may map onto the same points more than once, so the one-to-one condition fails. To apply the formula directly, use an angular interval that does not duplicate points in the interior. A shared boundary ray or other boundary overlap does not affect area because boundaries of this kind have area zero.

These conditions explain why the Jacobian formula is not simply a rule to attach to any parameter bounds: the map must represent the region without interior overlap. When it does, the absolute determinant supplies the correct local area scaling.

Common Mistake: Treating \(r\) as One Fixed Multiplier

A frequent error is to calculate the parameter area and multiply it by a single radius, as though the Jacobian were constant. For the region in the earlier example, the parameter area is \(\pi/3\). Using \(r=1\) as a supposed constant multiplier gives the incorrect result

$$ 1\cdot\frac{\pi}{3}=\frac{\pi}{3}, $$

which is not the image area. Using \(r=2\) would give \(2\pi/3\), also incorrect. The correction is to integrate the local factor throughout the region:

$$ \int_0^{\pi/3}\int_1^2 r\,dr\,d\theta=\frac{\pi}{2}. $$

Only when the Jacobian is constant over the entire parameter region can one multiply its area by a single scaling factor. In polar coordinates \(r\) generally varies, so the integral is essential.

Why the Polar Area Element Matters

The element \(r\,dr\,d\theta\) is the bridge between a description in polar parameters and area in the plane. The factor \(r\) captures the geometric stretching of angular changes: at distance \(r\), a small angle \(d\theta\) corresponds to an arc length approximately \(r\,d\theta\), while a small radial change has length \(dr\). Their product gives the same local area scale as the Jacobian derivation.

Remember the distinction: \(r\,dr\,d\theta\) describes local area, while the area of a whole mapped region is obtained by integrating that local element over its parameter region. This keeps the geometry and the calculation consistent whenever the Jacobian varies.

Check Your Understanding

Try these questions to check your understanding of the polar area element.

  1. Write the polar coordinate map and compute its Jacobian determinant.
  2. Why does the determinant depend on \(r\), and what does that dependence mean geometrically?
  3. For \(1\le r\le 3\) and \(0\le\theta\le\pi/2\), set up an integral for the image area. Why can you not replace \(r\) by one fixed value?
  4. What happens to the Jacobian determinant at \(r=0\), and why does this not prevent computing the area of a region that includes the origin?
  5. State the condition on the polar map that prevents counting an interior point of the image more than once.

Previous

Next

Back to Tutorials