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Chapter 15: Problem 4

In Exercises \(1-8,\) integrate the given function over the given surface. \(G(x, y, z)=z^{2},\) over the hemisphere \(x^{2}+y^{2}+\) \(z^{2}=a^{2}, z \geq 0\)

Short Answer

Expert verified
The integral of \(G(x, y, z) = z^{2}\) over the hemisphere is \(\frac{2\pi a^{4}}{3}.\)

Step by step solution

01

Understand the Surface

The given surface is a hemisphere defined by the equation \(x^{2}+y^{2}+z^{2} = a^{2}\) with \(z \geq 0\). This means we are dealing with the top half of a sphere with radius \(a\). In spherical coordinates, this surface can be parameterized with \(z = a\cos\theta\).
02

Convert to Spherical Coordinates

Convert the function \(G(x, y, z) = z^2\) to spherical coordinates. In these coordinates, \(z = a\cos\phi\), where \(\phi\) is the polar angle. Thus, \(G(x, y, z) = (a\cos\phi)^{2} = a^{2}\cos^{2}\phi\).
03

Write the Surface Integral

The surface integral of a function \(G(x, y, z)\) over a surface \(S\) is expressed as \(\iint_{S} G(x, y, z) \, dS\). In spherical coordinates, the differential area element \(dS\) for a hemisphere is \(a^{2} \sin\phi \, d\phi \, d\theta\).
04

Set the Limits of Integration

For the hemisphere, \(\phi\) varies from \(0\) to \(\pi/2\) (since we're considering the upper hemisphere where \(z \geq 0\)) and \(\theta\) varies from \(0\) to \(2\pi\).
05

Integrate over the Hemisphere

The integral becomes: \[\int_0^{2\pi} \int_0^{\pi/2} a^{2} \cos^{2}\phi \cdot a^{2} \sin\phi \, d\phi \, d\theta.\]This simplifies to: \[a^{4} \int_0^{2\pi} \int_0^{\pi/2} \cos^{2}\phi \sin\phi \, d\phi \, d\theta.\]
06

Perform Inner Integration

First integrate with respect to \(\phi\): \[\int_0^{\pi/2} \cos^{2}\phi \sin\phi \, d\phi.\]Substitute \(u = \cos\phi\), then \(du = -\sin\phi \, d\phi\), adjusting the limits, this integral simplifies to:\[-\int_1^0 u^{2} \, du = \left[\frac{-u^{3}}{3}\right]_{1}^0 = \frac{1}{3}.\]
07

Perform Outer Integration

Substituting back the result of the inner integral, we perform the outer integral with respect to \(\theta\): \[a^{4} \int_0^{2\pi} \frac{1}{3} \, d\theta = \frac{a^{4}}{3}\left[\theta\right]_0^{2\pi} = \frac{a^{4}}{3} \times 2\pi = \frac{2\pi a^{4}}{3}.\]

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Key Concepts

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

Spherical Coordinates
Spherical coordinates provide a way to describe points in three-dimensional space using three values: the radial distance, the polar angle, and the azimuthal angle. It serves as an alternative to the conventional Cartesian coordinates \((x, y, z)\). This system is beneficial for problems involving spherical symmetries, such as surface integrals over spheres and hemispheres. Here, the position of a point is given by \((r, \theta, \phi)\), where:
  • \(r\) is the radial distance from the origin to the point.
  • \(\theta\) is the azimuthal angle in the \(xy\)-plane from the positive \(x\)-axis.
  • \(\phi\) is the polar angle from the positive \(z\)-axis.
For a hemisphere, the relation \(z = r\cos\phi\) is particularly useful, helping to express functions like \(G(x, y, z) = z^2\) in spherical coordinates. This conversion highlights the use of spherical coordinates to simplify the integration over curved surfaces.
Hemispheres
The hemisphere in this context refers to half of a sphere. Specifically, it is defined as the upper half of a sphere when cut along a plane through its center. The equation \(x^2 + y^2 + z^2 = a^2\) and \(z \geq 0\) represents the upper hemisphere with radius \(a\) centered at the origin. This shape is relevant in many physical and geometric problems because of its symmetry and smooth surface.
The hemisphere can be parameterized effectively using spherical coordinates. By doing this, the surface becomes a set of points defined by the angles \(\theta\) and \(\phi\), helping in setting up limits for integration. Understanding hemispheres is crucial as they are often used in calculating properties like area and volume or solving physical equations over spherical surfaces.
Integration Limits
Integration limits in surface integrals define the range over which the integral is calculated. For surface integrals over a hemisphere, the range of angles \(\phi\) and \(\theta\) are essential. These limits ensure that the entire surface, and no more, is covered by the integration.
- The polar angle \(\phi\) varies from \(0\) to \(\pi/2\), corresponding to the upper hemisphere from the zenith (top) down to the equator.- The azimuthal angle \(\theta\) ranges from \(0\) to \(2\pi\), covering a full rotation around the \(z\)-axis.Setting these specific limits allows us to capture the hemisphere's full surface, making sure that calculations align precisely with the actual geometric space intended for the problem. This precision is vital for obtaining accurate results from surface integrals.
Parameterization of Surfaces
In mathematics, parameterization of a surface involves expressing the coordinates of every point on the surface in terms of two parameters. This method is particularly useful in converting complex surfaces into manageable forms for calculation purposes, such as integration.
For the hemisphere example, parameterization is done using spherical coordinates:
  • \( x = a \sin\phi \cos\theta \)
  • \( y = a \sin\phi \sin\theta \)
  • \( z = a \cos\phi \)
This conversion simplifies the surface integral by reducing it to an integral over the parameters \(\phi\) and \(\theta\). Parameterization transforms the conceptual complexity of a surface to a planar form, thereby utilizing the parameter space for efficient calculations. This is crucial when dealing with curved surfaces such as hemispheres.

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Most popular questions from this chapter

Calculating Area with Green's Theorem If a simple closed curve C in the plane and the region \(R\) it encloses satisfy the hypotheses of Green's Theorem, the area of \(R\) is given by Green's Theorem Area Formula Area of \(R=\frac{1}{2} \oint_{c} x d y-y d x\) The reason is that, by Equation (4) run backward, Area of \(R=\iint_{k} d y d x=\iint_{k}\left(\frac{1}{2}+\frac{1}{2}\right) d y d x\) $$=\oint_{c} \frac{1}{2} x d y-\frac{1}{2} y d x$$ Use the Green's Theorem area formula given above to find the areas of the regions enclosed by the curves in Exercises \(31-34\) The ellipse \(\mathbf{r}(t)=(a \cos t) \mathbf{i}+(b \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi\)

Bendixson's criterion The streamlines of a planar fluid flow are the smooth curves traced by the fluid's individual particles. The vectors \(\mathbf{F}=M(x, y) \mathbf{i}+N(x, y) \mathbf{j}\) of the flow's velocity field are the tangent vectors of the streamlines. Show that if the flow takes place over a simply connected region \(R\) (no holes or missing points \()\) and that if \(M_{x}+N_{y} \neq 0\) throughout \(R,\) then none of the streamlines in \(R\) is closed. In other words, no particle of fluid ever has a closed trajectory in \(R\). The criterion \(M_{x}+N_{y} \neq 0\) is called Bendixson's criterion for the nonexistence of closed trajectories.

In Exercises 29 and \(30 .\) find the surface integral of the field \(\mathbf{F}\) over the portion of the given surface in the specified direction. \(\mathbf{F}(x, y, z)=-\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}\) \(S:\) rectangular surface \(z=0, \quad 0 \leq x \leq 2, \quad 0 \leq y \leq 3\) direction k

Let \(C\) be the smooth curve \(\mathbf{r}(t)=(2 \cos t) \mathbf{i}+(2 \sin t) \mathbf{j}+ \left(3-2 \cos ^{3} t\right) \mathbf{k},\) oriented to be traversed counterclockwise around the \(z\) -axis when viewed from above. Let \(S\) be the piece wise smooth cylindrical surface \(x^{2}+y^{2}=4,\) below the curve for \(z \geq 0\) together with the base disk in the \(x y\) -plane. Note that \(C\) lies on the cylinder \(S\) and above the \(x y\) -plane (see the accompanying figure). Verify Equation (4) in Stokes' Theorem for the vector field \(\mathbf{F}=y \mathbf{i}-x \mathbf{j}+x^{2} \mathbf{k}.\) (Graph cant copy)

Find the area of the surface cut from the paraboloid \(x^{2}+y^{2}-z=0\) by the plane \(z=2\)
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