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Chapter 5: Problem 35

If \(\rho(x, y)\) is the density of a wire (mass per unit length), then the mass of the wire is \(m=\int_{C} \rho(x, y) d s\). Find the mass of a wire having the shape of a semicircle \(x=1+\cos t, y=\sin t, 0 \leq t \leq \pi\), if the density at a point \(P\) is directly proportional to the distance from the \(y\) -axis.

Short Answer

Expert verified
The mass of the wire is \(k\pi\).

Step by step solution

01

Understand the Problem

We need to find the mass of a wire shaped like a semicircle, described parametrically by the equations \(x = 1 + \cos t\) and \(y = \sin t\), where \(0 \leq t \leq \pi\). The density is directly proportional to the distance from the \(y\)-axis. Thus, the density \(\rho(x,y) = k \cdot |x|\), where \(k\) is the proportionality constant.
02

Express Density in Terms of Parameter t

Since the density \(\rho(x, y) = k \cdot |x|\) and \(x = 1+ \cos t\), the distance from the \(y\)-axis is given by the \(x\)-coordinate value. Therefore, \(\rho(x,y) = k \cdot |1 + \cos t|\). As \(0 \leq t \leq \pi\), \(\cos t\) ranges from 1 to -1, so \(|1 + \cos t|\) simplifies to \(1 + \cos t\). Hence, \(\rho(x, y) = k(1 + \cos t)\).
03

Calculate the Differential Arc Length d s

The element of arc length \(d s\) for a parametric curve is \(d s = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt\). Calculate \(\frac{dx}{dt} = -\sin t\) and \(\frac{dy}{dt} = \cos t\). Therefore, \(d s = \sqrt{(-\sin t)^2 + (\cos t)^2} dt = \sqrt{\sin^2 t + \cos^2 t} dt = dt\).
04

Set Up the Integral to Find the Mass

The mass of the wire is given by the integral \(m = \int_{C} \rho(x, y) d s = \int_{0}^{\pi} k(1 + \cos t) dt\).
05

Evaluate the Integral

Evaluate the integral \(m = \int_{0}^{\pi} k(1 + \cos t) dt = k\left[\int_{0}^{\pi} 1\, dt + \int_{0}^{\pi} \cos t\, dt \right]\). Integrate each part separately: \(\int_{0}^{\pi} 1\, dt = \pi\) and \(\int_{0}^{\pi} \cos t\, dt = [\sin t]_{0}^{\pi} = 0\). So, the integral becomes \(k[\pi + 0] = k\pi\).
06

Conclude with the Mass Expression

The mass of the wire is \(m = k\pi\), which shows that the mass depends linearly on the proportionality constant \(k\), which is determined by specific physical properties provided.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parameterization
In calculus, parameterization refers to representing a curve through equations that express the coordinates of the points on the curve as functions of a variable called a parameter. For example, in the given problem, we have a semicircle described by the parametric equations
  • \(x = 1 + \cos t\)
  • \(y = \sin t\)
where the parameter \(t\) ranges from 0 to \(\pi\). These equations allow us to rewrite the original geometric object (the semicircle) in a more flexible form. Parameterization helps us compute various properties of the curve, like its length or mass, more easily. For the semicircle, the parameter \(t\) dictates the position along the curve, providing a systematic way to analyze every point on the semicircle.
Density Function
A density function in the context of this problem refers to how mass is distributed along the wire. Here, the wire's density is directly proportional to the distance from the \(y\)-axis, expressed as \(\rho(x, y) = k \cdot |x|\), where \(k\) is the proportionality constant. In our parameterized semicircle path, the x-coordinate represents this distance:
  • Since \(x = 1 + \cos t\), the distance from the \(y\)-axis becomes \(|1 + \cos t|\).
  • This leads to \(\rho(x, y) = k(1 + \cos t)\), simplifying our work as it no longer involves dealing with the absolute value.
By using the parameter \(t\), we translate the difficulty of calculating the direct distance from axes into something more manageable in terms of integration.
Arc Length
The arc length of a parametric curve is calculated using the differentials of the parameterized equations. In our semicircle problem, the element of arc length \(d s\) is given by:\[d s = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt\]Substituting the derivatives:
  • \(\frac{dx}{dt} = -\sin t\)
  • \(\frac{dy}{dt} = \cos t\)
we get:\[d s = \sqrt{(-\sin t)^2 + (\cos t)^2} dt = \sqrt{\sin^2 t + \cos^2 t} dt = dt\]This results in a simplified form of arc length due to the identity \(\sin^2 t + \cos^2 t = 1\), making further calculations straightforward. The arc length calculation is essential for integrating along the entire curve to find properties like mass.
Integral Calculation
The integral calculation here is about finding the total mass along the semicircle of the wire by integrating the density function over the arc length. Our setup for the mass \(m\) becomes:\[m = \int_{C} \rho(x, y) d s = \int_{0}^{\pi} k(1 + \cos t) dt\]Breaking down this integral involves:
  • First calculating \(\int_{0}^{\pi} 1 \ dt = \pi\)
  • Then, separately \(\int_{0}^{\pi} \cos t \ dt = [\sin t]_{0}^{\pi} = 0\)
Thus, the total mass simplifies to \(m = k \pi\). This showcases the ingenuity of calculus in combining parameterization and integration to solve complex physical problems, ensuring we accurately calculate properties like mass distribution in non-linear shapes.

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Most popular questions from this chapter

Use any of the results in this section to evaluate the given integral along the indicated closed contour(s). \(\oint_{C} \frac{-3 z+2}{z^{2}-8 z+12} d z\) (a) \(|z-5|=2\), (b) \(|z|=9\)

For any two real numbers \(k\) and \(x_{1}\), the function \(\Omega(z)=k \operatorname{Ln}\left(z-x_{1}\right)\) is analytic in the upper half-plane and therefore is complex potential for the flow of an ideal fluid. The real number \(x_{1}\) is a sink when \(k<0\) and a source for the flow when \(k>0\) (a) Show that the streamlines are rays emanating from \(x_{1}\). (b) Show that the complex representation \(f(z)\) of the velocity field \(\mathbf{F}(x, y)\) of the flow is $$ f(z)=k \frac{z-x_{1}}{\left|z-x_{1}\right|^{2}} $$ and conclude that the flow is directed toward \(x_{1}\) precisely when \(k<0 .\)

Evaluate \(\int_{C}(2 x+y) d x+x y d y\) on the given curve from (-1,2) to (2,5). $$ y=x^{2}+1 $$

Evaluate \(\oint_{C}\left(x^{2}-y^{2}\right) d s\), where \(C\) is given by \(x=5 \cos t, y=5 \sin t, 0 \leq t \leq 2 \pi\).

Evaluate the given integral along the indicated contour. \(\int_{C} \sin z d z\), where \(C\) is the polygonal path consisting of the line segments from \(z=0\) to \(z=1\) and from \(z=1\) to \(z=1+i\)
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