Center of Mass for MATH 141
Exam Relevance for MATH 141
Center of mass problems appear on some MATH 141 exams. Know the integral formulas for 1D and 2D.
Understanding Center of Mass
If you've ever tried to balance a ruler on your finger, you've found its center of mass — the point where all the weight seems to be concentrated. For a uniform ruler, it's the middle. But what about irregular shapes or regions with varying density?
Calculus lets us find the center of mass for continuous objects and planar regions. The key idea: the center of mass is the "weighted average" position, where position is weighted by mass (or area).
Centroids of Planar Regions
For a region with uniform density, the center of mass is called the centroid. It depends only on the geometry, not the material.
Centroid Formulas
For a region bounded above by $y = f(x)$, below by the x-axis, from $x = a$ to $x = b$:
$$\bar{x} = \frac{1}{A} \int_a^b x \cdot f(x) \, dx$$
$$\bar{y} = \frac{1}{A} \int_a^b \frac{1}{2}[f(x)]^2 \, dx$$
Where $A = \int_a^b f(x) \, dx$ is the area of the region.
Why These Formulas?
- $\bar{x}$: Each vertical strip at position $x$ has area $f(x)\,dx$. We weight position by area.
- $\bar{y}$: Each strip extends from $0$ to $f(x)$, so its "middle" is at height $\frac{f(x)}{2}$. The $\frac{1}{2}[f(x)]^2$ comes from weighting this middle height by the strip's area.
Problem: Find the centroid of the region bounded by $y = x$, $y = 0$, and $x = 2$.
Step 1: Find the area
$$A = \int_0^2 x \, dx = \left[\frac{x^2}{2}\right]_0^2 = 2$$
Step 2: Find $\bar{x}$
$$\bar{x} = \frac{1}{2} \int_0^2 x \cdot x \, dx = \frac{1}{2} \int_0^2 x^2 \, dx$$
$$= \frac{1}{2} \left[\frac{x^3}{3}\right]_0^2 = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}$$
Step 3: Find $\bar{y}$
$$\bar{y} = \frac{1}{2} \int_0^2 \frac{1}{2}x^2 \, dx = \frac{1}{4} \int_0^2 x^2 \, dx$$
$$= \frac{1}{4} \cdot \frac{8}{3} = \frac{2}{3}$$
$$\boxed{(\bar{x}, \bar{y}) = \left(\frac{4}{3}, \frac{2}{3}\right)}$$
Problem: Find the centroid of the region bounded by $y = 4 - x^2$ and the x-axis.
Step 1: Find the bounds
$4 - x^2 = 0$ gives $x = \pm 2$.
Step 2: Find the area
$$A = \int_{-2}^{2} (4 - x^2) \, dx = \left[4x - \frac{x^3}{3}\right]_{-2}^{2}$$
$$= \left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right) = \frac{32}{3}$$
Step 3: Find $\bar{x}$
By symmetry, $\bar{x} = 0$ (the region is symmetric about the y-axis).
Step 4: Find $\bar{y}$
$$\bar{y} = \frac{1}{A} \int_{-2}^{2} \frac{1}{2}(4 - x^2)^2 \, dx$$
$$= \frac{3}{32} \cdot \frac{1}{2} \int_{-2}^{2} (16 - 8x^2 + x^4) \, dx$$
$$= \frac{3}{64} \left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_{-2}^{2}$$
At $x = 2$: $32 - \frac{64}{3} + \frac{32}{5} = \frac{256}{15}$
At $x = -2$: $-32 + \frac{64}{3} - \frac{32}{5} = -\frac{256}{15}$
$$\bar{y} = \frac{3}{64} \cdot \frac{512}{15} = \frac{8}{5}$$
$$\boxed{(\bar{x}, \bar{y}) = \left(0, \frac{8}{5}\right)}$$
Centroid Between Two Curves
For a region between $y = f(x)$ (top) and $y = g(x)$ (bottom):
$$\bar{x} = \frac{1}{A} \int_a^b x[f(x) - g(x)] \, dx$$
$$\bar{y} = \frac{1}{A} \int_a^b \frac{1}{2}\left([f(x)]^2 - [g(x)]^2\right) dx$$
Where $A = \int_a^b [f(x) - g(x)] \, dx$.
Problem: Find the centroid of the region bounded by $y = x^2$ and $y = x$.
Step 1: Find intersection points
$x^2 = x$ gives $x = 0$ and $x = 1$.
Step 2: Find the area
$$A = \int_0^1 (x - x^2) \, dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$$
Step 3: Find $\bar{x}$
$$\bar{x} = \frac{1}{1/6} \int_0^1 x(x - x^2) \, dx = 6 \int_0^1 (x^2 - x^3) \, dx$$
$$= 6 \left[\frac{x^3}{3} - \frac{x^4}{4}\right]_0^1 = 6\left(\frac{1}{3} - \frac{1}{4}\right) = 6 \cdot \frac{1}{12} = \frac{1}{2}$$
Step 4: Find $\bar{y}$
$$\bar{y} = 6 \int_0^1 \frac{1}{2}(x^2 - x^4) \, dx = 3 \int_0^1 (x^2 - x^4) \, dx$$
$$= 3 \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = 3\left(\frac{1}{3} - \frac{1}{5}\right) = 3 \cdot \frac{2}{15} = \frac{2}{5}$$
$$\boxed{(\bar{x}, \bar{y}) = \left(\frac{1}{2}, \frac{2}{5}\right)}$$
Center of Mass with Variable Density
When density varies as $\rho(x)$, the formulas become:
$$\bar{x} = \frac{\int_a^b x \cdot \rho(x) \cdot f(x) \, dx}{\int_a^b \rho(x) \cdot f(x) \, dx}$$
The denominator is the total mass $M$, and the numerator is the "moment" about the y-axis.
Problem: A rod extends from $x = 0$ to $x = 3$ with density $\rho(x) = x^2$ kg/m. Find its center of mass.
Step 1: Find total mass
$$M = \int_0^3 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^3 = 9 \text{ kg}$$
Step 2: Find the moment
$$M_y = \int_0^3 x \cdot x^2 \, dx = \int_0^3 x^3 \, dx = \left[\frac{x^4}{4}\right]_0^3 = \frac{81}{4}$$
Step 3: Calculate center of mass
$$\bar{x} = \frac{M_y}{M} = \frac{81/4}{9} = \frac{81}{36} = \frac{9}{4} = 2.25$$
$$\boxed{\bar{x} = \frac{9}{4} = 2.25 \text{ m}}$$
The center of mass is closer to $x = 3$ because the rod is denser there.
Problem: Find the centroid of the region in the first quadrant bounded by $x^2 + y^2 = r^2$.
Step 1: Set up
The boundary is $y = \sqrt{r^2 - x^2}$ from $x = 0$ to $x = r$.
Step 2: Find the area
$$A = \frac{1}{4}\pi r^2$$ (quarter of a circle)
Step 3: Find $\bar{x}$
By symmetry with $\bar{y}$ (quarter circle is symmetric about $y = x$), we have $\bar{x} = \bar{y}$.
$$\bar{x} = \frac{4}{\pi r^2} \int_0^r x\sqrt{r^2 - x^2} \, dx$$
Let $u = r^2 - x^2$, so $du = -2x\,dx$.
$$= \frac{4}{\pi r^2} \cdot \left(-\frac{1}{2}\right) \cdot \frac{2}{3}u^{3/2}\Big|_{r^2}^{0}$$
$$= \frac{4}{\pi r^2} \cdot \frac{1}{3} \cdot r^3 = \frac{4r}{3\pi}$$
$$\boxed{(\bar{x}, \bar{y}) = \left(\frac{4r}{3\pi}, \frac{4r}{3\pi}\right)}$$
Pappus's Theorem
Bonus insight: The volume of a solid of revolution equals the area of the region times the distance traveled by its centroid:
$$V = 2\pi \bar{r} \cdot A$$
where $\bar{r}$ is the distance from the centroid to the axis of rotation.
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting to divide by the area
Wrong: $\bar{x} = \int_a^b x \cdot f(x) \, dx$
Why it's wrong: This gives the moment, not the center of mass. You need to divide by total area (or mass).
Correct: $\bar{x} = \frac{1}{A} \int_a^b x \cdot f(x) \, dx$
❌ Mistake: Using $f(x)$ instead of $\frac{1}{2}[f(x)]^2$ for $\bar{y}$
Wrong: $\bar{y} = \frac{1}{A} \int_a^b f(x) \, dx$
Why it's wrong: That would give you the average value of $f$, not the y-coordinate of the centroid. The $\bar{y}$ formula accounts for the "middle height" of each strip.
Correct: $\bar{y} = \frac{1}{A} \int_a^b \frac{1}{2}[f(x)]^2 \, dx$
❌ Mistake: Ignoring symmetry
Wrong: Computing $\bar{x}$ with a long integral when the region is symmetric about the y-axis.
Why it's wrong: You're doing unnecessary work! Symmetry tells you the answer immediately.
Correct: If a region is symmetric about the y-axis, then $\bar{x} = 0$. Use symmetry to save time.
❌ Mistake: Using the wrong formula for regions between curves
Wrong: Using $\frac{1}{2}[f(x)]^2$ for a region that doesn't start at $y = 0$.
Why it's wrong: The basic formula assumes the region is bounded below by the x-axis.
Correct: For regions between $f(x)$ and $g(x)$, use $\frac{1}{2}([f(x)]^2 - [g(x)]^2)$ for $\bar{y}$.
Centroid x-coordinate
The x-coordinate of the centroid for a region under y = f(x). Divide the moment about the y-axis by the total area.
Variables:
- $x̄$:
- x-coordinate of centroid
- $A$:
- area of the region
- $f(x)$:
- the curve bounding the region above
Centroid y-coordinate
The y-coordinate of the centroid for a region under y = f(x). The ½[f(x)]² accounts for the middle height of each vertical strip.
Variables:
- $ȳ$:
- y-coordinate of centroid
- $A$:
- area of the region
- $f(x)$:
- the curve bounding the region above
Center of Mass (1D with variable density)
Center of mass for a one-dimensional object with density ρ(x). The numerator is the moment; the denominator is total mass.
Variables:
- $x̄$:
- center of mass position
- $ρ(x)$:
- density function
- $a, b$:
- endpoints of the object
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