Setting Up Lamina Integrals With Horizontal Slices

Why Use Horizontal Slices?

A lamina is a thin, flat object whose mass is spread over a two-dimensional region. To calculate its mass and first moments, we can divide the region into thin strips and add the contributions from those strips. With horizontal slices, each strip lies at an approximately fixed \(y\)-coordinate and extends from a left boundary to a right boundary.

This description is especially useful when the region’s left and right edges can be written naturally as functions of \(y\). Each horizontal strip then has the same \(y\)-value throughout, while its \(x\)-values range between the two boundary curves.

Definition: Suppose a region \(R\) is described by \(c\le y\le d\) and \(L(y)\le x\le R(y)\), where \(L(y)\) and \(R(y)\) are its left and right boundaries. A horizontal slice at height \(y\) runs from \(x=L(y)\) to \(x=R(y)\). Its thickness is \(dy\).

Here, \(R(y)\) denotes the right-boundary function; \(R\) without an argument denotes the entire planar region. We assume that the boundaries are well-defined on the stated interval, with \(L(y)\le R(y)\). If the region has different left or right boundaries over different \(y\)-intervals, describe and integrate those pieces separately.

Mass and First Moments from a Horizontal Strip

Let \(\rho(x,y)\) be the areal density of the lamina: mass per unit area at the point \((x,y)\). Assume the density is nonnegative and continuous, or piecewise continuous so the integrals exist. A small area element has area \(dA=dx\,dy\), so its mass is \(dm=\rho(x,y)\,dx\,dy\).

The mass is the sum of the masses of all the area elements. The first moment about the \(x\)-axis weights each mass element by its vertical distance \(y\); the first moment about the \(y\)-axis weights it by its horizontal distance \(x\). Therefore, a region described by horizontal slices has these iterated-integral formulas:

m=\int_c^d\int_{L(y)}^{R(y)}\rho(x,y)\,dx\,dy

M_x=\int_c^d\int_{L(y)}^{R(y)}y\,\rho(x,y)\,dx\,dy, \qquad M_y=\int_c^d\int_{L(y)}^{R(y)}x\,\rho(x,y)\,dx\,dy.

The inner integral accounts for the mass or moment across one horizontal strip, and the outer integral adds those contributions as \(y\) runs from \(c\) to \(d\). Notice the different coordinate factors: \(M_x\) contains \(y\), while \(M_y\) contains \(x\).

Why These Integrals Work

The formulas follow by adding the contributions of increasingly thin pieces. Divide the vertical interval from \(c\) to \(d\) into short intervals of width \(\Delta y\). In one such interval near height \(y_i\), the horizontal strip runs approximately from \(L(y_i)\) to \(R(y_i)\). Divide that strip into small widths \(\Delta x\).

At a sample point \((x_{ij},y_i)\), the small rectangular piece has area approximately \(\Delta x\,\Delta y\), so its mass is approximately \(\rho(x_{ij},y_i)\Delta x\,\Delta y\). For its contribution to \(M_x\), multiply by its \(y\)-coordinate; for \(M_y\), multiply by its \(x\)-coordinate. Adding all these approximations gives sums of the form below. As the widths tend to zero, the sums become the iterated integrals.

\begin{aligned} m&=\lim\sum_{i,j}\rho(x_{ij},y_i)\,\Delta x\,\Delta y,\\ M_x&=\lim\sum_{i,j}y_i\rho(x_{ij},y_i)\,\Delta x\,\Delta y,\\ M_y&=\lim\sum_{i,j}x_{ij}\rho(x_{ij},y_i)\,\Delta x\,\Delta y. \end{aligned}

This reasoning also explains why a variable density must remain inside the integrals: different parts of a strip can have different densities. Replacing \(\rho(x,y)\) by one constant is justified only when the density really is constant throughout the lamina.

Useful Formulas for Constant Density

If the density is a constant \(\rho_0\), the inner integrals can be evaluated in terms of the strip’s boundaries. The strip has width \(R(y)-L(y)\), so its area is approximately \([R(y)-L(y)]\,dy\). Its mass is \(\rho_0\) times that area.

Takeaway: For constant areal density \(\rho_0\),

\begin{aligned} m&=\rho_0\int_c^d\bigl(R(y)-L(y)\bigr)\,dy,\\ M_x&=\rho_0\int_c^d y\bigl(R(y)-L(y)\bigr)\,dy,\\ M_y&=\frac{\rho_0}{2}\int_c^d\bigl(R(y)^2-L(y)^2\bigr)\,dy. \end{aligned}

The last formula comes from integrating \(x\) across each strip: \(\int_{L(y)}^{R(y)}x\,dx=[R(y)^2-L(y)^2]/2\).

Worked Examples

Worked Example: A Uniform Rectangle

A rectangular lamina occupies \(-1\le y\le 1\) and \(-2\le x\le 1\), with constant density \(\rho_0=2\) in the chosen mass-per-area units. Find its mass and both first moments using horizontal slices.

For every \(y\) in the interval, the left boundary is \(-2\) and the right boundary is \(1\). The mass is

\begin{aligned} m &=2\int_{-1}^{1}\int_{-2}^{1}1\,dx\,dy\\ &=2\int_{-1}^{1}[x]_{-2}^{1}\,dy\\ &=2\int_{-1}^{1}3\,dy\\ &=2(3)(2)=12. \end{aligned}

For the moment about the \(x\)-axis, the inner integral is the strip’s mass before multiplication by \(y\):

\begin{aligned} M_x &=2\int_{-1}^{1}y\int_{-2}^{1}1\,dx\,dy\\ &=6\int_{-1}^{1}y\,dy\\ &=6\left[\frac{y^2}{2}\right]_{-1}^{1}=0. \end{aligned}

For the moment about the \(y\)-axis, integrate \(x\) across the strip:

\begin{aligned} M_y &=2\int_{-1}^{1}\int_{-2}^{1}x\,dx\,dy\\ &=2\int_{-1}^{1}\left[\frac{x^2}{2}\right]_{-2}^{1}\,dy\\ &=2\int_{-1}^{1}\left(\frac12-2\right)\,dy\\ &=2\left(-\frac32\right)(2)=-6. \end{aligned}

Check the arithmetic in the inner integral: \(\frac12-2=-\frac32\), and the interval of \(y\)-values has length \(2\). Thus \(M_y=-6\). The antiderivatives used above differentiate to \(1\), \(y\), and \(x\), respectively, as required by the integrands.

Worked Example: A Shifted Region with Variable Density

Let the region be \(0\le y\le 1\), \(y\le x\le y+1\), and let its density be \(\rho(x,y)=1+x+y\). Compute the mass and both first moments. This example shows why the horizontal boundaries and the density must both be used in the integrals.

For the mass, integrate the density across a strip first. With \(x=y+t\), where \(0\le t\le1\), the density becomes \(1+2y+t\). Hence

\begin{aligned} m &=\int_0^1\int_y^{y+1}(1+x+y)\,dx\,dy\\ &=\int_0^1\int_0^1(1+2y+t)\,dt\,dy\\ &=\int_0^1\left(\frac32+2y\right)\,dy\\ &=\left[\frac32y+y^2\right]_0^1=\frac52. \end{aligned}

For \(M_x\), the factor \(y\) is constant across each horizontal strip, so it can be taken outside the inner integral:

\begin{aligned} M_x &=\int_0^1 y\int_y^{y+1}(1+x+y)\,dx\,dy\\ &=\int_0^1 y\left(\frac32+2y\right)\,dy\\ &=\left[\frac34y^2+\frac23y^3\right]_0^1\\ &=\frac34+\frac23=\frac{17}{12}. \end{aligned}

For \(M_y\), retain the factor \(x\) inside the integral across the strip. Using \(x=y+t\) again gives

\begin{aligned} \int_y^{y+1}x(1+x+y)\,dx &=\int_0^1(y+t)(1+2y+t)\,dt\\ &=\int_0^1\bigl(y+2y^2+(1+3y)t+t^2\bigr)\,dt\\ &=y+2y^2+\frac{1+3y}{2}+\frac13\\ &=2y^2+\frac52y+\frac56. \end{aligned}

Therefore,

\begin{aligned} M_y &=\int_0^1\left(2y^2+\frac52y+\frac56\right)\,dy\\ &=\left[\frac23y^3+\frac54y^2+\frac56y\right]_0^1\\ &=\frac23+\frac54+\frac56=\frac{33}{12}=\frac{11}{4}. \end{aligned}

As checks, differentiating the mass antiderivative \(\frac32y+y^2\) gives \(\frac32+2y\), and differentiating the outer antiderivatives for the moments gives their respective integrands \(\frac32y+2y^2\) and \(2y^2+\frac52y+\frac56\). The final arithmetic is exact.

Worked Example: A Parabolic Boundary

Consider the region \(0\le y\le1\), \(y^2\le x\le1\), with constant density \(\rho_0=1\). Set up and evaluate the mass and first moments. The left boundary changes with height, so the strip width is \(1-y^2\).

\begin{aligned} m &=\int_0^1\int_{y^2}^{1}1\,dx\,dy\\ &=\int_0^1(1-y^2)\,dy\\ &=\left[y-\frac{y^3}{3}\right]_0^1=\frac23. \end{aligned}

The first moment about the \(x\)-axis is

\begin{aligned} M_x &=\int_0^1\int_{y^2}^{1}y\,dx\,dy\\ &=\int_0^1y(1-y^2)\,dy\\ &=\left[\frac{y^2}{2}-\frac{y^4}{4}\right]_0^1 =\frac12-\frac14=\frac14. \end{aligned}

The first moment about the \(y\)-axis requires integrating \(x\) across each strip:

\begin{aligned} M_y &=\int_0^1\int_{y^2}^{1}x\,dx\,dy\\ &=\int_0^1\left[\frac{x^2}{2}\right]_{y^2}^{1}\,dy\\ &=\frac12\int_0^1(1-y^4)\,dy\\ &=\frac12\left[y-\frac{y^5}{5}\right]_0^1 =\frac12\left(1-\frac15\right)=\frac25. \end{aligned}

Each result can be checked by differentiating its displayed outer antiderivative: the derivatives are \(1-y^2\), \(y-y^3\), and \(\frac12(1-y^4)\), which are the corresponding outer integrands.

A Common Setup Error

A frequent mistake is to use limits from the wrong description of the region. For the shifted region in the second example, each horizontal strip runs from \(x=y\) to \(x=y+1\). Replacing those limits by \(0\) and \(1\) treats the region as a square instead. For the same density, that incorrect setup gives

\int_0^1\int_0^1(1+x+y)\,dx\,dy =\int_0^1\left(\frac32+y\right)\,dy=2,

not the correct mass \(\frac52\). The error is not in the integration: it is in describing a different region. Before integrating, check that at each height \(y\), the inner \(x\)-limits actually run from the left edge of the lamina to its right edge.

A related mistake is to interchange the coordinate factors in the first moments. The distance to the \(x\)-axis is \(|y|\), so the signed first-moment integrand uses \(y\rho(x,y)\); the distance to the \(y\)-axis is \(|x|\), so its integrand uses \(x\rho(x,y)\). These signed coordinates allow contributions on opposite sides of an axis to cancel.

Setting Up Reliably

Before evaluating any integrals, identify the vertical range of the region and determine the left and right \(x\)-values at a typical height. Then write the mass and moment integrands using the appropriate coordinate factors. This separates the geometric task—describing the region—from the integration task.

Describe the vertical range.
Write \(c\le y\le d\), splitting the range if the boundary description changes.

Find the horizontal endpoints.
For each height, identify \(L(y)\) on the left and \(R(y)\) on the right.

Insert density and moment factors.
Use \(\rho\) for mass, \(y\rho\) for \(M_x\), and \(x\rho\) for \(M_y\).

Integrate across and then through the region.
Integrate with respect to \(x\) from \(L(y)\) to \(R(y)\), then with respect to \(y\) from \(c\) to \(d\).

In physical applications, if \(x\) and \(y\) measure length and \(\rho\) has units of mass per area, \(m\) has units of mass and each first moment has units of mass times length. Keeping these units in mind can help detect a missing density or coordinate factor. Horizontal-slice integrals thus provide one consistent way to model a lamina, whether its density is uniform or varies from point to point.

Check Your Understanding

For each question, identify the region description or integral structure that makes the setup correct.

For a region with \(2\le y\le5\), left boundary \(L(y)\), and right boundary \(R(y)\), write the mass integral for density \(\rho(x,y)\).
Which coordinate factor belongs in the integrand for the first moment about the \(x\)-axis?
For constant density \(\rho_0\), what is the horizontal strip’s contribution to \(M_y\) per unit \(y\)-thickness?
In a horizontal-slice description, what should be checked if the left boundary changes at some height?
Why is it generally incorrect to pull a variable density \(\rho(x,y)\) outside the inner integral?