Rewriting Integrands in Polar Coordinates

Why Rewrite an Integrand?

When a region is naturally described using distance from the origin and angle, Cartesian expressions in \(x\) and \(y\) can obscure useful structure. Rewriting the function in terms of \(r\) and \(\theta\) can reveal that structure and make a polar double integral easier to set up or evaluate.

The substitution itself is direct: replace \(x\) by \(r\cos\theta\) and \(y\) by \(r\sin\theta\). But there are two distinct changes to keep track of. The function changes because its inputs are being expressed in polar coordinates; the area element changes because a small polar region has a different area scale. These should be handled separately.

Key distinction: Rewriting a function \(f(x,y)\) gives \(f(r\cos\theta,r\sin\theta)\). In a polar double integral, the area element contributes an additional factor \(r\), so the complete expression is \(f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta\).

The Substitution and the Area Factor

For \(r\ge 0\), polar coordinates are related to Cartesian coordinates by

x=r\cos\theta, \qquad y=r\sin\theta.

Thus, for a function \(f\) of two variables, its expression in polar coordinates is obtained by direct substitution:

f(x,y)\longmapsto f(r\cos\theta,r\sin\theta).

This gives the transformed function. It does not yet include the area factor. The polar area element is \(dA=r\,dr\,d\theta\). Consequently, when a double integral over a region \(D\) is expressed over its corresponding polar parameter region \(R\), the change-of-variables formula is

\iint_D f(x,y)\,dA = \iint_R f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta.

Here \(R\) describes the radial and angular values for the region. Determining those limits is a separate part of setting up the integral. In this tutorial, the emphasis is on rewriting the function and keeping the area factor distinct.

The factor \(r\) comes from the area scaling of the polar coordinate map. Its Jacobian determinant is

\frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta+r\sin^2\theta = r.

Since \(r\ge 0\), the absolute value of this determinant is \(r\). It scales the area element; it is not part of the original function \(f\). This is why the substitution and the area factor should be written as separate steps.

A Useful Algebraic Rule

Applying the substitution to a product of powers gives a convenient rule. If \(m\) and \(n\) are nonnegative integers, then

x^m y^n \longmapsto (r\cos\theta)^m(r\sin\theta)^n = r^{m+n}\cos^m\theta\sin^n\theta.

The rule follows by replacing each occurrence of \(x\) and \(y\), then collecting the powers of \(r\), \(\cos\theta\), and \(\sin\theta\). In particular, a monomial of total degree \(m+n\) contributes \(r^{m+n}\) to the transformed function. The additional \(r\) from the area element is still applied afterward.

Takeaway: First transform every \(x\) and \(y\) in the function. Then, when writing the polar double integral, multiply the transformed function by the area factor \(r\) exactly once.

Worked Examples

Worked Example: A Quadratic Polynomial

Rewrite \(f(x,y)=3x^2+4y^2-2xy\) in polar coordinates, then include the area factor.

Substitute \(x=r\cos\theta\) and \(y=r\sin\theta\) into each term:

\begin{aligned} f(r\cos\theta,r\sin\theta) &=3(r\cos\theta)^2+4(r\sin\theta)^2 -2(r\cos\theta)(r\sin\theta)\\ &=3r^2\cos^2\theta+4r^2\sin^2\theta -2r^2\cos\theta\sin\theta\\ &=r^2\left(3\cos^2\theta+4\sin^2\theta -2\cos\theta\sin\theta\right). \end{aligned}

This is the transformed function. Multiplying it by the area factor gives the expression in the polar integral:

f(r\cos\theta,r\sin\theta)\,r = r^3\left(3\cos^2\theta+4\sin^2\theta -2\cos\theta\sin\theta\right).

To check the transformed function, use \(r^2\cos^2\theta=x^2\), \(r^2\sin^2\theta=y^2\), and \(r^2\cos\theta\sin\theta=xy\). The expression becomes \(3x^2+4y^2-2xy\), as required. The extra \(r\) belongs to the area element, not to this check of the function.

Worked Example: A Function of \(x^2+y^2\)

Rewrite \(g(x,y)=(x^2+y^2)e^{-(x^2+y^2)}\) and then include the area factor.

Substitute into the sum of squares first:

\begin{aligned} x^2+y^2 &=(r\cos\theta)^2+(r\sin\theta)^2\\ &=r^2\cos^2\theta+r^2\sin^2\theta\\ &=r^2(\cos^2\theta+\sin^2\theta)\\ &=r^2. \end{aligned}

Therefore, the transformed function is

g(r\cos\theta,r\sin\theta) = r^2 e^{-r^2}.

Including the polar area factor gives

g(r\cos\theta,r\sin\theta)\,r = r^3e^{-r^2}.

For a check, substitute \(r^2=x^2+y^2\) back into \(r^2e^{-r^2}\). The result is \((x^2+y^2)e^{-(x^2+y^2)}\), which is the original function. Again, the check concerns the transformed function alone; the factor \(r\) is added for area.

Worked Example: A Square Root and a Product

Rewrite \(h(x,y)=\sqrt{x^2+y^2}+xy\) in polar coordinates and include the area factor. Since \(r\ge 0\), the square root simplifies without an absolute-value sign:

\begin{aligned} \sqrt{x^2+y^2} &=\sqrt{r^2\cos^2\theta+r^2\sin^2\theta}\\ &=\sqrt{r^2(\cos^2\theta+\sin^2\theta)}\\ &=\sqrt{r^2}\\ &=r. \end{aligned}

The product becomes

xy=(r\cos\theta)(r\sin\theta) =r^2\cos\theta\sin\theta.

Combining these terms, the transformed function is

h(r\cos\theta,r\sin\theta) = r+r^2\cos\theta\sin\theta.

The expression including the area factor is

h(r\cos\theta,r\sin\theta)\,r = r^2+r^3\cos\theta\sin\theta.

To check, use \(r=\sqrt{x^2+y^2}\) and \(r^2\cos\theta\sin\theta=xy\). The transformed function returns to \(\sqrt{x^2+y^2}+xy\). The area-weighted expression is not meant to return to \(h(x,y)\), because it includes the separate area scaling factor.

Why the Substitution Works

The replacement is not a special rule for particular expressions. It follows from composing the original function with the coordinate map. The map sends a polar pair \((r,\theta)\) to the Cartesian point \((r\cos\theta,r\sin\theta)\). Evaluating \(f\) at that point therefore gives \(f(r\cos\theta,r\sin\theta)\). The area element then accounts for how the map scales small areas.

For example, for a monomial \(x^m y^n\), repeated multiplication gives

\begin{aligned} x^m y^n &=(r\cos\theta)^m(r\sin\theta)^n\\ &=r^m\cos^m\theta\,r^n\sin^n\theta\\ &=r^{m+n}\cos^m\theta\sin^n\theta. \end{aligned}

This proves the stated power rule. When the monomial is used in a polar double integral, the full factor multiplying \(dr\,d\theta\) is then

r^{m+n+1}\cos^m\theta\sin^n\theta.

The exponent increases by one because of the area factor, not because the function itself changed again.

A Common Mistake: Counting the Area Factor Twice

A frequent error is to include \(r\) while rewriting the function and then multiply by another \(r\) for the area element. Take \(f(x,y)=x^2+y^2\). The function substitution is

f(r\cos\theta,r\sin\theta) = r^2\cos^2\theta+r^2\sin^2\theta = r^2.

The correct factor in the polar integral is \(r^2\cdot r=r^3\). A mistaken calculation might label \(r^3\) as the transformed function, then multiply by \(r\) again and use \(r^4\). That is wrong: \(r^3\) already includes the area factor. Checking the roles separately prevents the error: the function becomes \(r^2\), and the area element contributes one \(r\), giving \(r^3\).

Another useful check is to ask whether the expression being written is the function or the full factor in the integral. State that distinction explicitly during setup, especially when the original function already contains powers of \(x^2+y^2\), which themselves become powers of \(r\).

When This Rewriting Is Useful

Expressions built from \(x^2+y^2\) often simplify because that sum becomes \(r^2\). Products and powers become combinations of powers of \(r\), \(\cos\theta\), and \(\sin\theta\). Such forms can make radial dependence and angular dependence easier to recognize. They also help prevent a common conceptual mix-up: the transformed function describes the original quantity at a point, while the area factor accounts for the size of the polar area element around that point.

For a complete polar integral, both pieces are necessary. Omitting the area factor changes the area weighting; including it twice changes the integral as well. The region's radial and angular limits must also match the region being integrated over, but those limits do not alter the substitution rule for the function.

Check Your Understanding

For each question, distinguish the transformed function from the factor that includes the polar area element.

Rewrite \(x^2+3y^2\) in polar coordinates. What is the expression after multiplying by the area factor?
What does \(xy^2\) become under the substitution?
If \(f(x,y)=e^{x^2+y^2}\), what is \(f(r\cos\theta,r\sin\theta)\)?
For \(f(x,y)=x^2+y^2\), explain why the polar integral has a factor \(r^3\), not \(r^2\) or \(r^4\).
How can you check that a proposed transformed function is correct?