Integrating Below an Inclined Plane in Cylindrical Coordinates

Introduction

An inclined plane does not usually have the same height at every point above a disk or sector. Its height changes with position, so an integral beneath it must account for both the distance from the axis and the angle around the axis.

Cylindrical coordinates make that variation explicit. They also make the intersection with a vertical cylinder easy to describe: on the cylindrical wall, the radius is fixed, while the plane’s height may still vary with angle.

Writing a Plane as a Cylindrical Height

A plane can be written in Cartesian form \(Ax+By+Cz=D\). If \(C\ne 0\), it can be solved for \(z\), so it is the graph of a height function over the \(xy\)-plane. Using \(x=r\cos\theta\) and \(y=r\sin\theta\), write that function as

z=h(r,\theta)=c+ar\cos\theta+br\sin\theta,

where \(a,b,c\) are constants. The coefficients can be positive or negative. The condition \(C\ne0\) matters: if \(C=0\), the plane is vertical and cannot be represented as a single height \(z=h(x,y)\) over its projection.

Definition: For a region \(D\) in the \(r\)-\(\theta\) plane, the solid beneath the plane and above \(z=0\) is \(0\le z\le h(r,\theta)\) for \((r,\theta)\in D\), provided \(h(r,\theta)\ge0\) throughout \(D\).

In cylindrical coordinates, the plane’s height on the side of a cylinder of radius \(R\) is found by setting \(r=R\):

z=h(R,\theta)=c+aR\cos\theta+bR\sin\theta.

This gives the plane-cylinder intersection as a curve on the cylinder’s side. The curve is generally not at a constant height: its height changes as \(\theta\) changes.

Checking Whether the Plane Stays Above the Base

Before integrating from \(z=0\) to the plane, check that the plane does not dip below the base on the chosen projection. For a full disk \(0\le r\le R\), the expression \(a\cos\theta+b\sin\theta\) ranges between \(-\sqrt{a^2+b^2}\) and \(\sqrt{a^2+b^2}\). Consequently, the minimum height over the disk is

c-R\sqrt{a^2+b^2}.

Thus, \(c\ge R\sqrt{a^2+b^2}\) ensures nonnegative height everywhere on the disk. If the inequality is strict, the plane is strictly above the base. If it fails, the plane may cross \(z=0\) inside the projection; integrating from zero to a negative height would not describe the intended solid. The projection or the vertical bounds must then be adjusted.

Setting Up the Triple Integral

Suppose the projected region is described by \(\alpha\le\theta\le\beta\) and \(r_{\min}(\theta)\le r\le r_{\max}(\theta)\), and suppose \(h\ge0\) there. The cylindrical volume element is \(r\,dz\,dr\,d\theta\). Therefore the volume beneath the plane is

V=\int_{\alpha}^{\beta} \int_{r_{\min}(\theta)}^{r_{\max}(\theta)} \int_0^{h(r,\theta)} r\,dz\,dr\,d\theta.

The factor \(r\) is the cylindrical-coordinate scale factor. Integrating first with respect to \(z\) replaces the inner integral by the height \(h(r,\theta)\), leaving an integral over the projection.

Key takeaway: The upper vertical bound is the plane’s height \(h(r,\theta)\), including its angular dependence. On a cylindrical wall, substitute the wall radius into that same height function.

Why the Height Integral Gives Volume

At a fixed point \((r,\theta)\) in the projection, the vertical segment in the solid runs from \(z=0\) to \(z=h(r,\theta)\). Integrating \(dz\) over that segment gives its length. The cylindrical volume element multiplies that length by the appropriate area contribution \(r\,dr\,d\theta\).

\int_0^{h(r,\theta)} r\,dz = r\left[z\right]_0^{h(r,\theta)} = r h(r,\theta).

Integrating these contributions over the entire projected region adds the volumes of all the vertical slices. This explains both the upper bound and why the factor \(r\) remains after the \(z\)-integration.

Worked Examples

Worked Example: A Tilted Plane over a Full Cylinder

Find the volume inside \(r\le2\), above \(z=0\), and below the plane \(z=6+x-2y\). First rewrite the height in cylindrical coordinates:

h(r,\theta)=6+r\cos\theta-2r\sin\theta.

Over the disk, the minimum height is \(6-2\sqrt{1^2+(-2)^2}=6-2\sqrt5>0\), so the full disk is a valid projection. On the cylinder wall \(r=2\), the plane has height \(6+2\cos\theta-4\sin\theta\).

Set up and evaluate the volume integral:

\begin{aligned} V &=\int_0^{2\pi}\int_0^2\int_0^{6+r\cos\theta-2r\sin\theta}r\,dz\,dr\,d\theta\\ &=\int_0^{2\pi}\int_0^2(6+r\cos\theta-2r\sin\theta)r\,dr\,d\theta\\ &=\int_0^{2\pi}\left(12+\frac{8}{3}\cos\theta-\frac{16}{3}\sin\theta\right)d\theta\\ &=24\pi. \end{aligned}

For the radial integration, \(\int_0^2 6r\,dr=12\) and \(\int_0^2r^2\,dr=8/3\). Over a full turn, both \(\int_0^{2\pi}\cos\theta\,d\theta\) and \(\int_0^{2\pi}\sin\theta\,d\theta\) equal zero. As a check, the constant height \(6\) contributes \(6\cdot\pi(2)^2=24\pi\); the two linear terms average to zero over the full disk, agreeing with the integral.

Worked Example: A Plane over a Quarter-Cylinder

Find the volume inside \(0\le r\le2\), \(0\le\theta\le\pi/2\), above the base and below \(z=5-x-y\). The cylindrical height and the height on the outer wall are

h(r,\theta)=5-r(\cos\theta+\sin\theta), \qquad h(2,\theta)=5-2(\cos\theta+\sin\theta).

Because \(\cos\theta+\sin\theta\le\sqrt2\), the height is at least \(5-2\sqrt2>0\) over this quarter-disk. Now integrate:

\begin{aligned} V &=\int_0^{\pi/2}\int_0^2\int_0^{5-r(\cos\theta+\sin\theta)}r\,dz\,dr\,d\theta\\ &=\int_0^{\pi/2}\int_0^2\left(5r-r^2(\cos\theta+\sin\theta)\right)dr\,d\theta\\ &=\int_0^{\pi/2}\left(10-\frac{8}{3}(\cos\theta+\sin\theta)\right)d\theta\\ &=5\pi-\frac{16}{3}. \end{aligned}

Here \(\int_0^2 5r\,dr=10\), \(\int_0^2r^2\,dr=8/3\), and each of \(\int_0^{\pi/2}\cos\theta\,d\theta\) and \(\int_0^{\pi/2}\sin\theta\,d\theta\) equals \(1\). As a check, the integrand for the projected region is \(h(r,\theta)r\), which is nonnegative because \(h\ge5-2\sqrt2>0\) and \(r\ge0\); the resulting volume is positive.

Worked Example: An Inclined Plane over a Narrow Sector

Find the volume inside \(0\le r\le3\), \(-\pi/6\le\theta\le\pi/6\), above \(z=0\), and below \(z=4-\tfrac12x\). In cylindrical coordinates,

h(r,\theta)=4-\frac12r\cos\theta.

Since \(\cos\theta\le1\) on this sector and \(r\le3\), the height is at least \(4-\tfrac32=\tfrac52>0\). At the cylindrical wall, the height is \(4-\tfrac32\cos\theta\). Integrating gives

\begin{aligned} V &=\int_{-\pi/6}^{\pi/6}\int_0^3\int_0^{4-\frac12r\cos\theta}r\,dz\,dr\,d\theta\\ &=\int_{-\pi/6}^{\pi/6}\int_0^3\left(4r-\frac12r^2\cos\theta\right)dr\,d\theta\\ &=\int_{-\pi/6}^{\pi/6}\left(18-\frac92\cos\theta\right)d\theta\\ &=6\pi-\frac92. \end{aligned}

Indeed, \(\int_0^3 4r\,dr=18\), \(\int_0^3r^2\,dr=9\), and \(\int_{-\pi/6}^{\pi/6}\cos\theta\,d\theta=2\sin(\pi/6)=1\). The computed volume is positive, as required for a solid with positive height throughout its projection.

A Common Mistake: Dropping the Angular Dependence

For the plane \(z=5-x-y\), it is incorrect to write the cylindrical height as \(5-r\). That would assume \(x+y=r\), which is not generally true. The correct conversion is \(x+y=r\cos\theta+r\sin\theta\), so the height is \(5-r(\cos\theta+\sin\theta)\).

For example, at \(r=1\) and \(\theta=\pi\), the point has \(x=-1\) and \(y=0\). The actual plane height is \(5-(-1)-0=6\). The incorrect expression \(5-r\) gives \(4\), which does not match the plane. Substituting both Cartesian coordinates before simplifying prevents this error.

Using the Method Reliably

For a solid bounded above by a plane and laterally by a cylinder, identify the projection and its radial and angular bounds first. Then express the plane as \(h(r,\theta)\), check that the height is nonnegative on the chosen projection, and evaluate the triple integral with \(z\) running from the base to that height. Evaluating the height at the cylinder radius describes where the plane meets the side; it does not replace the height function throughout the interior.

Solve the plane for height.
Use \(C\ne0\) to express \(z\) as a function of \(x\) and \(y\).

Convert to cylindrical coordinates.
Substitute \(x=r\cos\theta\) and \(y=r\sin\theta\).

Check the boundary height and positivity.
Set \(r=R\) for the cylinder wall and verify that the height is nonnegative over the projection.

Integrate the slices.
Use the height as the upper \(z\)-bound and include the cylindrical factor \(r\).

Check Your Understanding

Use the height function and the projected region to answer these questions.

Rewrite \(z=7+2x-y\) in cylindrical coordinates.
For a cylinder of radius \(3\), what is the plane’s height on its side?
What condition on \(c,a,b,\) and \(R\) guarantees that \(c+ar\cos\theta+br\sin\theta\) is nonnegative throughout a full disk of radius \(R\)?
Set up, but do not evaluate, the volume integral beneath \(z=4-x\) over the quarter-disk \(0\le r\le1\), \(0\le\theta\le\pi/2\), above \(z=0\).
Why is \(z=5-r\) not the correct cylindrical form of \(z=5-x-y\)?