Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Discrete mathematics, continuous mathematics and probability / 5 / c

Table of contents
- Solution c

Solution

A horizontal strip at height

y

has width

4 - 2 sin (π y /2)

. The sine term is odd, so the area is

∣ A ∣ = \int_{- 2}^{2} (4 - 2 sin \frac{π y}{2}) d y = 16.

Reflection symmetry in the

y

-axis gives

\overset{x}{ˉ} = 0

. For uniform density, the constant density cancels from the center of mass ratio. Its first moment in the vertical coordinate is

\int_{A} y d A = - 4 \int_{0}^{2} y sin \frac{π y}{2} d y = - 4 [- \frac{2 y}{π} cos \frac{π y}{2} + \frac{4}{π ^{2}} sin \frac{π y}{2}]_{0}^{2} = - \frac{16}{π} .

Consequently

(\overset{x}{ˉ}, \overset{y}{ˉ}) = (0, - \frac{1}{π}) .

The wider part below the horizontal axis accounts for the negative vertical coordinate.

Figure 1.
Uniform-density plate with area 16 and centroid at (0, -1/π)
. Original plot of the strip-defined region. The marked centroid lies below the horizontal axis; the highlighted sample strip illustrates the width used in the area and moment integrals. Details.

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