Rewriting Cartesian Integrands in Cylindrical Form

Why Rewrite the Integrand?

Cylindrical coordinates are useful when a solid or its bounds are naturally described using distance from the vertical axis and angle around that axis. To express a triple integral in these coordinates, the integrand must also be rewritten using the cylindrical variables. The key is to convert the Cartesian expression itself carefully before combining it with the volume element.

The basic substitutions are \(x=r\cos\theta\) and \(y=r\sin\theta\). They immediately give \(x^2+y^2=r^2\), but products and powers require attention: for example, \(xy\) contains two factors of \(r\), not one. Keeping track of those factors prevents many conversion errors.

Definition: For cylindrical coordinates, \(r\ge 0\) is the distance from the \(z\)-axis, \(\theta\) is the angle in the \(xy\)-plane, and \(z\) is unchanged. The Cartesian and cylindrical coordinates are related by \(x=r\cos\theta\), \(y=r\sin\theta\), and \(z=z\).

The Basic Conversion Rules

Substitute the coordinate formulas into each part of the Cartesian integrand. The most frequently used identities follow directly:

x=r\cos\theta,\qquad y=r\sin\theta,\qquad x^2+y^2=r^2.

For the last identity, square the first two formulas and add:

x^2+y^2 =r^2\cos^2\theta+r^2\sin^2\theta =r^2(\cos^2\theta+\sin^2\theta) =r^2.

A product such as \(xy\) must be converted by substituting for both factors:

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

More generally, for nonnegative integers \(m\) and \(n\),

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

This follows by substituting \(r\cos\theta\) for each of the \(m\) factors of \(x\), and \(r\sin\theta\) for each of the \(n\) factors of \(y\). Thus the total power of \(r\) is \(m+n\). Expressions that depend only on \(x^2+y^2\) are radial: replacing \(x^2+y^2\) by \(r^2\) removes their angular dependence.

Important distinction: Rewriting the Cartesian function gives \(f(r\cos\theta,r\sin\theta,z)\). In a cylindrical triple integral, this function is then multiplied by the volume-element factor \(r\). The factor from the volume element is separate from the powers of \(r\) already present in the rewritten function.

A Derivation of the Substitution

The conversion rules are not separate tricks for particular expressions. They follow from the coordinate map itself. In the \(xy\)-plane, a point at distance \(r\) and angle \(\theta\) has horizontal coordinate \(r\cos\theta\) and vertical coordinate \(r\sin\theta\). The height coordinate is unaffected. Therefore, if a Cartesian function is \(f(x,y,z)\), its expression in cylindrical coordinates is obtained by composing \(f\) with this map.

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

For example, the identity for \(x^2+y^2\) is a consequence of the Pythagorean identity \(\cos^2\theta+\sin^2\theta=1\), as shown above. Likewise, replacing both factors in \(xy\) gives \(r^2\cos\theta\sin\theta\). These substitutions preserve the value of the function at the same point in space: the Cartesian and cylindrical coordinates name that point in different ways.

The cylindrical volume element is \(r\,dz\,dr\,d\theta\). It accounts for the change in volume when coordinates change; it is not part of the coordinate substitution for the function. Consequently, the full integrand-and-volume-element expression is

f(r\cos\theta,r\sin\theta,z)\,r\,dz\,dr\,d\theta.

This explains why it is useful to do the conversion in two explicit stages: first rewrite the function, then include the volume element.

Worked Conversions

Worked Example: Rewriting a Polynomial

Rewrite \(3x^2+2xy+y^2+4z\) in cylindrical coordinates.

Substitute for each occurrence of \(x\) and \(y\), leaving \(z\) unchanged:

\begin{aligned} 3x^2+2xy+y^2+4z &=3(r\cos\theta)^2 +2(r\cos\theta)(r\sin\theta) +(r\sin\theta)^2+4z\\ &=3r^2\cos^2\theta +2r^2\cos\theta\sin\theta +r^2\sin^2\theta+4z. \end{aligned}

The first and third terms can also be grouped, but they should not be combined as though their coefficients were equal:

r^2(3\cos^2\theta+\sin^2\theta) +2r^2\cos\theta\sin\theta+4z.

To check the result, substitute \(x=r\cos\theta\) and \(y=r\sin\theta\) into the original Cartesian expression. Its three terms involving \(x\) and \(y\) become \(3r^2\cos^2\theta\), \(2r^2\cos\theta\sin\theta\), and \(r^2\sin^2\theta\), respectively, while \(4z\) remains \(4z\). These are exactly the terms in the converted expression.

Worked Example: A Product Times a Radial Power

Rewrite \(xy(x^2+y^2)^2\) in cylindrical coordinates, and then include the cylindrical volume-element factor.

First convert the product:

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

Next convert the radial power:

$$ (x^2+y^2)^2=(r^2)^2=r^4. $$

Multiplying these converted factors gives the function:

xy(x^2+y^2)^2 =(r^2\cos\theta\sin\theta)(r^4) =r^6\cos\theta\sin\theta.

The volume element supplies one additional factor of \(r\), so the full expression in the integral is

r^6\cos\theta\sin\theta\;r\,dz\,dr\,d\theta =r^7\cos\theta\sin\theta\,dz\,dr\,d\theta.

For a check, substituting \(x=r\cos\theta\), \(y=r\sin\theta\), and \(x^2+y^2=r^2\) into the Cartesian function gives \(r^2\cos\theta\sin\theta\cdot r^4\), which is \(r^6\cos\theta\sin\theta\). The extra power producing \(r^7\) comes only from the volume element.

Worked Example: A Difference of Radial Powers

Rewrite \(\dfrac{z}{x^2+y^2}+(x^2+y^2)^{3/2}\) in cylindrical coordinates and state where the expression is defined.

The denominator is nonzero only when \(x^2+y^2\ne 0\), which corresponds to \(r\ne 0\). On that domain, convert each term:

\frac{z}{x^2+y^2} =\frac{z}{r^2}, \qquad (x^2+y^2)^{3/2} =(r^2)^{3/2} =r^3,

where \(r\ge 0\) is used in \((r^2)^{3/2}=r^3\). Thus the converted function is

\frac{z}{r^2}+r^3,\qquad r>0.

Including the volume element gives

\left(\frac{z}{r^2}+r^3\right)r\,dz\,dr\,d\theta = \left(\frac{z}{r}+r^4\right)dz\,dr\,d\theta, \qquad r>0.

To check the function conversion, replace \(r^2\) by \(x^2+y^2\) in the first term and \(r^3\) by \((x^2+y^2)^{3/2}\) in the second. This recovers the original function. The restriction \(r>0\) also matches the original denominator's restriction \(x^2+y^2\ne0\).

Products, Sums, and Powers

A reliable conversion can be done term by term. First identify each occurrence of \(x\), \(y\), and \(x^2+y^2\). Substitute the coordinate formulas, simplify powers of \(r\), and only then collect terms if useful. For instance, a sum such as \(x^2+y^2+xy\) becomes

r^2+r^2\cos\theta\sin\theta =r^2(1+\cos\theta\sin\theta).

The first two squares combine into \(r^2\), while the product contributes its own angular factor. This is a useful distinction: not every expression involving \(x\) and \(y\) is radial. Dependence only on \(x^2+y^2\) is radial, but a term such as \(xy\) generally depends on \(\theta\) as well as \(r\).

Powers also require attention to parentheses. The expressions \((x^2+y^2)^2\) and \(x^2+y^2\) become \(r^4\) and \(r^2\), respectively. The exponent applies to the entire quantity \(x^2+y^2\), so replacing that quantity by \(r^2\) must happen before applying the outer power.

A Common Conversion Mistake

A frequent error is to assign only one factor of \(r\) to \(xy\). The incorrect calculation is

xy=(r\cos\theta)(r\sin\theta) \overset{\text{incorrect}}{=}r\cos\theta\sin\theta.

That line has dropped one of the two factors of \(r\). Multiplying the factors correctly gives

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

One way to catch the error is to count the powers: \(x\) contributes one \(r\), and \(y\) contributes another. A second check is to substitute the correct expression back into Cartesian coordinates: \(r^2\cos\theta\sin\theta=(r\cos\theta)(r\sin\theta)=xy\). The incorrect expression would instead correspond to \(xy/r\) when \(r>0\).

Putting the Conversion Into an Integral

Once the integrand has been converted, it can be placed in the cylindrical integral with the bounds appropriate to the solid. The conversion procedure does not itself determine those bounds; it supplies the expression to use with bounds already written in cylindrical coordinates. For example, if the Cartesian integrand is \(x^2+y^2+z\), then

$$ x^2+y^2+z=r^2+z, $$

and the corresponding integrand with the volume element is

(r^2+z)\,r\,dz\,dr\,d\theta.

The expression \(r^2+z\) is the converted function. The full factor \( (r^2+z)r \) is what multiplies the differentials in the triple integral. Keeping these two descriptions distinct makes it easier to check both the algebra and the coordinate-change factor.

Why Careful Conversion Matters

The integrand records the quantity being accumulated over the solid. A conversion that loses a factor of \(r\), changes an angular factor, or alters the domain of a function changes that quantity. Before proceeding to integration, check that every Cartesian variable has been replaced, that \(z\) has been preserved, and that any restrictions such as a nonzero denominator have been translated correctly.

Substitute.
Replace each \(x\) by \(r\cos\theta\), each \(y\) by \(r\sin\theta\), and leave \(z\) unchanged.

Simplify.
Use \(x^2+y^2=r^2\), count powers of \(r\) in products, and simplify the angular factors.

Check the domain.
Carry over restrictions such as a denominator being nonzero.

Attach the volume element.
Multiply the converted function by \(r\,dz\,dr\,d\theta\) to form the cylindrical integral.

Check Your Understanding

For each question, keep the converted function distinct from the cylindrical volume element.

Rewrite \(x^2+4y^2\) in cylindrical coordinates.
Rewrite \(x^2y\), showing how many factors of \(r\) occur.
Rewrite \(z(x^2+y^2)^2\), and then include the volume-element factor.
For which values of \(r\) is \(\dfrac{1}{x^2+y^2}\) defined after conversion?
Explain why converting \(xy\) gives \(r^2\cos\theta\sin\theta\), not \(r\cos\theta\sin\theta\).