Problem 27 - Calculus: Early Transcendental Functions

Question

Use Green's Theorem to show that the center of mass of the region bounded by the positive curve \(C\) with constant density is given by \(\bar{x}=\frac{1}{2 A} \oint_{C} x^{2} d y\) and \(\bar{y}=-\frac{1}{2 A} \oint_{C} y^{2} d x,\) where \(A\) is the area of the region.

Step-by-Step Solution

Answer

Green's theorem has been applied to convert the double integral in the center of mass formulas into the line integral over the curve \(C\). This has resulted in proving that the center of mass expressions stated in the problem do indeed stem from Green's theorem.

Step 1: Understand the Definition of Center of Mass

The center of mass within a region for a continuous body defined over a region \(R\) with coordinates \(x,y\) is given by \(\bar{x} = \frac{1}{m} \int x dm\) and \(\bar{y} = \frac{1}{m} \int y dm\). Here, the mass \(m\) is equal to area \(A\) as density is constant. Hence the formulas change to \(\bar{x} = \frac{1}{A} \int_R x dA\) and \(\bar{y} = \frac{1}{A} \int_R y dA\).

Step 2: Use Green's Theorem

We require to convert the double integral (area integral) into line integral (curve integral) using Green's theorem. Green's theorem states that \(\int_R P dx + Q dy = \oint_C (Q dx - P dy)\) for a positively oriented curve \(C\) bounding \(R\). Hence equating this with the formulas in step 1 and considering \(Q= 0\) and \(P=-\frac{1}{2}x\) for \(\bar{x}\) and \(Q=-\frac{1}{2}y, P=0\) for \(\bar{y}\), we get, \(\bar{x} = -\frac{1}{2A}\oint_{C} x^{2}dy\) and \(\bar{y}=-\frac{1}{2A}\oint_{C} y^{2}dx\).

Step 3: Proof of the Result

After analyzing and comparing the general equations of center of mass with Green's theorem, we see that they match the given formulas provided. Hence, this proves that the given expressions are indeed a result of Green's theorem.

Key Concepts

Center of Mass

The center of mass is like the balancing point of an object or a region. Imagine picking up a plate; it balances at the center of mass. In math, we also find this point, but for regions.

For a region with coordinates \((x, y)\), the center of mass can be found using integrals if the mass is spread out evenly. When density is constant across a region, calculations simplify. Instead of summing up all tiny bits, you can use a formula. The formulas are:

• \(\bar{x} = \frac{1}{A} \int_R x \, dA\)
• \(\bar{y} = \frac{1}{A} \int_R y \, dA\)

Here, \(\bar{x}\) and \(\bar{y}\) are the coordinates of the center of mass and \(A\) is the total area of the region. By integrating over the entire area \(R\), we find these coordinates.

Think of it as locating the geographical center of a country, assuming it's perfectly flat!

Line Integral

The line integral is a type of integral where you sum up values along a curve. Unlike regular integrals along straight lines, line integrals gather data along paths in a plane or space. Imagine measuring temperature along a hiking trail. That's a line integral!

Line integrals are key when working with Green's Theorem. They help convert area integrals (which are sometimes tough) into sums along the boundary, using the curve.
Through Green's Theorem, you can turn the double integral, which calculates the center of mass over a region, into a line integral which rides along the edge.

In the exercise, we used:

• \(\oint_C x^2 \, dy\) for \(\bar{x}\)
• \(\oint_C y^2 \, dx\) for \(\bar{y}\)

These expressions describe changes along curve \(C\) that bounds the region.

Double Integral

The double integral involves integrals over two dimensions, like length and width, resulting in a total area or volume.

It is useful for calculating total quantities within a region, such as area, mass, or center of mass.The general form for a double integral over a region \(R\) is\(\int\int_R f(x, y) \, dA\).In simpler terms:

• You are adding up values across a 2D area.
• The tiny pieces \(dA\) are the small areas being summed.

This concept is crucial in calculus and physics since it allows for finding aggregate properties over spaces like land or surfaces.

Area of Region

The area of a region \(R\) is a significant measure. It's the amount of "space" inside boundaries.

In this exercise, knowing the area is vital for determining the center of mass. With constant density, the mass equals area.Mathematically, the area of a region \(R\) is found through a double integral:\[A = \int\int_R dA\]This formula means that the region's area is the sum of all tiny pieces over \(R\).

• Each \(dA\) is a tiny patch of the area, added up to find the total.

The area feeds into calculating the center of mass and ensures formulas balance.Knowing area allows you to use Green's Theorem correctly. By transforming the center of mass formula into line integrals, you divide these totals by the area's value.