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Chapter 11: Problem 19

Evaluate the surface integral \(\int \mathbf{r} \cdot d \mathbf{S}\), where \(\mathbf{r}\) is the position vector, over that part of the surface \(z=a^{2}-x^{2}-y^{2}\) for which \(z \geq 0\), by each of the following methods: (a) parameterize the surface as \(x=a \sin \theta \cos \phi, y=a \sin \theta \sin \phi, z=a^{2} \cos ^{2} \theta\), and show that $$ \mathbf{r} \cdot d \mathbf{S}=a^{4}\left(2 \sin ^{3} \theta \cos \theta+\cos ^{3} \theta \sin \theta\right) d \theta d \phi $$ (b) apply the divergence theorem to the volume bounded by the surface and the plane \(z=0\)

Short Answer

Expert verified
Using parameterization, \( \mathbf{r} \cdot d \mathbf{S} = a^{4} \left( 2 \sin^{3} \theta \cos \theta + \cos^{3} \theta \sin \theta \right) d \theta d \phi \). Using the divergence theorem, the integral evaluates to \(2 \pi a^{3}\).

Step by step solution

01

Parameterize the Surface

Parameterize the surface using spherical coordinates where \(x = a \sin \theta \cos \phi, y = a \sin \theta \sin \phi, z = a^{2} \cos^{2} \theta\). This parameterization covers the region of the surface given by \(z \geq 0\).
02

Compute the Position Vector \( \mathbf{r} \)

The position vector can be expressed as \( \mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \). Substitute the parameterization into this vector to get: \( \mathbf{r} = a \sin \theta \cos \phi \, \mathbf{i} + a \sin \theta \sin \phi \, \mathbf{j} + a^{2} \cos^{2} \theta \, \mathbf{k} \).
03

Find the Differential Surface Element \( d \mathbf{S} \)

To find \( d \mathbf{S} \), we need the cross product of the partial derivatives of \( \mathbf{r} \) with respect to \( \theta \) and \( \phi \): \( \mathbf{r}_{\theta} = \frac{\partial \mathbf{r}}{\partial \theta} = a \cos \theta \cos \phi \, \mathbf{i} + a \cos \theta \sin \phi \, \mathbf{j} - 2a^{2} \cos \theta \sin \theta \, \mathbf{k} \) and \( \mathbf{r}_{\phi} = \frac{\partial \mathbf{r}}{\partial \phi} = -a \sin \theta \sin \phi \, \mathbf{i} + a \sin \theta \cos \phi \, \mathbf{j} \).
04

Calculate the Cross Product

Calculate the cross product \( \mathbf{r}_{\theta} \times \mathbf{r}_{\phi} \): \[ \mathbf{r}_{\theta} \times \mathbf{r}_{\phi} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a \cos \theta \cos \phi & a \cos \theta \sin \phi & - 2a^{2} \cos \theta \sin \theta \ - a \sin \theta \sin \phi & a \sin \theta \cos \phi & 0 \end{array} \right| = 2a^{3} \sin^{2} \theta \cos \theta \, \mathbf{i} + 2a^{3} \sin \theta \cos^{2} \theta \, \mathbf{j} + a^{2} \sin \theta (\sin^{2} \theta + \cos^{2} \theta) \, \mathbf{k} \]
05

Compute \( \mathbf{r} \cdot d \mathbf{S} \)

The differential surface element is given by \( d \mathbf{S} = \mathbf{r}_{\theta} \times \mathbf{r}_{\phi} d \theta d \phi \). The dot product of \( \mathbf{r} \) and \( d \mathbf{S} \) is: \( \mathbf{r} \cdot d \mathbf{S} = (a \sin \theta \cos \phi \, \mathbf{i} + a \sin \theta \sin \phi \, \mathbf{j} + a^{2} \cos^{2} \theta \, \mathbf{k}) \cdot (2a^{3} \sin^{2} \theta \cos \theta \, \mathbf{i} + 2a^{3} \sin \theta \cos^{2} \theta \, \mathbf{j} + a^{2} \sin \theta (\sin^{2} \theta + \cos^{2} \theta) \, \mathbf{k}) \). Simplifying, \( \mathbf{r} \cdot d \mathbf{S} = a^{4} \left( 2 \sin^{3} \theta \cos \theta + \cos^{3} \theta \sin \theta \right) d \theta d \phi \).
06

Apply the Divergence Theorem

Using the divergence theorem, convert the surface integral \( \int \mathbf{r} \cdot d \mathbf{S} \) into a volume integral: \( \int_{V} abla \cdot \mathbf{r} \, dV \) where \( abla \cdot \mathbf{r} = 3 \), since \( abla \cdot \mathbf{r} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} \). So, \( \int_{V} 3 \, dV \).
07

Calculate the Volume Integral

The volume bounded by the surface and the plane \(z = 0\) is a hemisphere of radius \(a\). The volume of this region is \(\frac{1}{2} \times \frac{4}{3} \pi a^{3} = \frac{2 \pi a^{3}}{3} \). Thus, \( \int_{V} 3 \, dV = 3 \times \frac{2 \pi a^{3}}{3} = 2 \pi a^{3} \).

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Key Concepts

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

Parameterization
Parameterizing a surface is a way to represent the surface using variables that span its entire domain. To parameterize the given surface, we utilize spherical coordinates due to the symmetry of the paraboloid surface.

In spherical coordinates, the surface can be expressed as follows:
  • x = a \sin(\theta) \cos(\phi),
  • y = a \sin(\theta) \sin(\phi),
  • z = a^2 \cos^2(\theta).

Here, \(\theta\) and \(\phi\) are the parameters. This approach makes it easier to compute integrals over the surface since we can now use \(\theta\) and \(\phi\) to describe every point on the surface.
Cross Product
The cross product is essential in computing the differential surface element, \(d\mathbf{S}\). This element is derived by taking the cross product of the partial derivatives of the parameterized position vector \(\mathbf{r}(\theta, \phi)\).

The position vector is given by:
\(\mathbf{r} = a \sin(\theta) \cos(\phi) \mathbf{i} + a \sin(\theta) \sin(\phi) \mathbf{j} + a^2 \cos^2(\theta) \mathbf{k}\).
We compute the partial derivatives:
\(\mathbf{r}_{\theta} = a \cos(\theta) \cos(\phi) \mathbf{i} + a \cos(\theta) \sin(\phi) \mathbf{j} - 2a^2 \cos(\theta) \sin(\theta) \mathbf{k}\) and \(\mathbf{r}_{\phi} = -a \sin(\theta) \sin(\phi) \mathbf{i} + a \sin(\theta) \cos(\phi) \mathbf{j}\).

Now, the cross product \(\mathbf{r}_{\theta} \times \mathbf{r}_{\phi}\) yields:\[\mathbf{r}_{\theta} \times \mathbf{r}_{\phi} = 2a^3 \sin^2(\theta) \cos(\theta) \mathbf{i} + 2a^3 \sin(\theta) \cos^2(\theta) \mathbf{j} + a^2 \sin(\theta)(\sin^2(\theta) + \cos^2(\theta)) \mathbf{k}\]This cross product gives the direction and magnitude of the differential surface element, necessary for evaluating surface integrals.
Divergence Theorem
The divergence theorem is a powerful tool for converting a surface integral into a volume integral. It's especially useful when evaluating integrals over complex surfaces. The theorem states:
\[ \int_\Sigma \mathbf{F} \cdot d \mathbf{S} = \int_V abla \cdot \mathbf{F} \, dV \] where \(\Sigma\) is the surface enclosing the volume \(V\), and \(\mathbf{F}\) is a vector field.

In our problem, \(\mathbf{r}\) is the position vector field. So, \(abla \cdot \mathbf{r}\) is calculated as:
\(abla \cdot \mathbf{r} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 3\).

Applying the theorem, the surface integral \(\int \mathbf{r} \cdot d \mathbf{S}\) converts to:
\(\int_V 3 \, dV\).
Volume Integrals
Volume integrals involve integrating a function over a three-dimensional region. Here, we need to integrate \(3\) over the volume of the hemisphere bounded by \(z = a^2 - x^2 - y^2\) and the plane \(z = 0\).

The volume of a hemisphere of radius \(a\) is readily found and given by:
\(\frac{1}{2} \times \frac{4}{3} \pi a^3 = \frac{2\pi a^3}{3}\). Thus,

\[\int_V 3 \, dV = 3 \times \frac{2\pi a^3}{3} = 2 \pi a^3\] In this step, we compute the total volume integral which represents the sum of all the small cubes covering the volume under the given paraboloid. The outcome, \(2\pi a^3\), matches our expectation from the divergence theorem.

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Most popular questions from this chapter

Show that the expression below is equal to the solid angle subtended by a rectangular aperture of sides \(2 a\) and \(2 b\) at a point a distance \(c\) from the aperture along the normal to its centre: $$ \Omega=4 \int_{0}^{b} \frac{a c}{\left(y^{2}+c^{2}\right)\left(y^{2}+c^{2}+a^{2}\right)^{1 / 2}} d y $$ By setting \(y=\left(a^{2}+c^{2}\right)^{1 / 2} \tan \phi\), change this integral into the form $$ \int_{0}^{\phi_{1}} \frac{4 a c \cos \phi}{c^{2}+a^{2} \sin ^{2} \phi} d \phi $$ where \(\tan \phi_{1}=b /\left(a^{2}+c^{2}\right)^{1 / 2}\), and hence show that $$ \Omega=4 \tan ^{-1}\left[\frac{a b}{c\left(a^{2}+b^{2}+c^{2}\right)^{1 / 2}}\right] $$

Obtain an expression for the value \(\phi_{P}\) at a point \(P\) of a scalar function \(\phi\) that satisfies \(\nabla^{2} \phi=0\) in terms of its value and normal derivative on a surface \(S\) that encloses it, by proceeding as follows. (a) In Green's second theorem take \(\psi\) at any particular point \(Q\) as \(1 / r\), where \(r\) is the distance of \(Q\) from \(P\). Show that \(\nabla^{2} \varphi=0\) except at \(r=0\) (b) Apply the result to the doubly connected region bounded by \(S\) and a small sphere \(\Sigma\) of radius \(\delta\) centred on \(\mathrm{P}\). (c) Apply the divergence theorem to show that the surface integral over \(\Sigma\) involving \(1 / \delta\) vanishes, and prove that the term involving \(1 / \delta^{2}\) has the value \(4 \pi \phi_{P}\) (d) Conclude that $$ \phi p=-\frac{1}{4 \pi} \int_{S} \phi \frac{\partial}{\partial n}\left(\frac{1}{r}\right) d S+\frac{1}{4 \pi} \int_{S} \frac{1}{r} \frac{\partial \phi}{\partial n} d S $$ This important result shows that the value at a point \(P\) of a function \(\phi\) that satisfies \(\nabla^{2} \phi=0\) everywhere within a closed surface \(S\) that encloses \(P\) may be expressed entirely in terms of its value and normal derivative on \(S .\) This matter is taken up more generally in connection with Green's functions in chapter 19 and in connection with functions of a complex variable in section \(20.12\)

By making an appropriate choice for the functions \(P(x, y)\) and \(Q(x, y)\) that appear in Green's theorem in a plane, show that the integral of \(x-y\) over the upper half of the unit circle centred on the origin has the value \(-\frac{2}{3}\). Show the same result by direct integration in Cartesian coordinates.

A rigid body of volume \(V\) and surface \(S\) rotates with angular velocity \(\omega .\) Show that $$ \omega=-\frac{1}{2 V} \oint_{S} \mathbf{u} \times d \mathbf{S} $$ where \(\mathbf{u}(\mathbf{x})\) is the velocity of the point \(\mathbf{x}\) on the surface \(S\).

A vector field a is given by \(\left(z^{2}+2 x y\right) \mathbf{i}+\left(x^{2}+2 y z\right) \mathbf{j}+\left(y^{2}+2 z x\right) \mathbf{k}\). Show that \(\mathbf{a}\) is conservative and that the line integral \(\int \mathbf{a} \cdot d \mathbf{r}\) along any line joining \((1,1,1)\) and \((1,2,2)\) has the value 11 .
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