Chapter 16: Problem 15
\(5-18\) Evaluate the surface integral. $$\iint_{S}\left(x^{2} z+y^{2} z\right) d S$$ \(S\) is the hemisphere \(x^{2}+y^{2}+z^{2}=4, z \geqslant 0\)
Short Answer
Expert verified
The surface integral evaluates to \(16\pi\).
Step by step solution
01
Identify the Surface
The surface given is a hemisphere described by the equation \(x^2 + y^2 + z^2 = 4\) with \(z \geq 0\). This is the upper half of a sphere with radius 2.
02
Parametrize the Surface
Using spherical coordinates, the parameterization of the hemisphere is \((x, y, z) = (2 \sin \theta \cos \phi, 2 \sin \theta \sin \phi, 2 \cos \theta)\), where \(0 \leq \theta \leq \frac{\pi}{2}\) and \(0 \leq \phi \leq 2\pi\).
03
Calculate the Surface Element
For spherical coordinates, the infinitesimal surface element \(dS\) is given by \(R^2 \sin \theta\, d\theta \, d\phi\). Since \(R = 2\), we have \(dS = 4 \sin \theta\, d\theta \, d\phi\).
04
Substitute into the Integral
Substitute the parameterization into the function: \(x^2 z + y^2 z = (4 \sin^2 \theta \cos^2 \phi + 4 \sin^2 \theta \sin^2 \phi)(2 \cos \theta) = 8 \sin^2 \theta \cos \theta\).
05
Set Up the Integral
The integral over the hemisphere becomes \(\int_0^{2\pi} \int_0^{\frac{\pi}{2}} 8 \sin^2 \theta \cos \theta \cdot 4 \sin \theta \, d\theta \, d\phi\). Simplify to \(32 \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \sin^3 \theta \cos \theta \, d\theta \, d\phi\).
06
Compute the Angular Integral
First, compute \(\int_0^{\frac{\pi}{2}} \sin^3 \theta \cos \theta \, d\theta\) using substitution \(u = \sin \theta\), \(du = \cos \theta \, d\theta\). The limits change from \(0\) to \(0\) and \(\pi/2\) to \(1\), resulting in \(\int_0^1 u^3 \, du = \frac{1}{4}\).
07
Integrate Over \(\phi\)
The integral over \(\phi\) is \(\int_0^{2\pi} d\phi = 2\pi\).
08
Evaluate the Integral
Combine the results: \(32 \times \frac{1}{4} \times 2\pi = 16\pi\).
09
Conclusion: Final Answer
The value of the surface integral \(\iint_{S}(x^{2} z+y^{2} z) \, dS\) over the hemisphere is \(16\pi\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Spherical Coordinates
Spherical coordinates are a system of coordinates that describe a point in three-dimensional space with three numbers: radius, polar angle, and azimuthal angle. This system is particularly useful when dealing with surfaces like spheres or hemispheres.
- **Radius** \(R\): Distance from the origin.- **Polar Angle** (\(\theta\)): Angle from the positive z-axis.- **Azimuthal Angle** (\(\phi\)): Angle in the xy-plane from the positive x-axis.
In the context of our given hemisphere, the radius \(R\) is 2, the polar angle \(\theta\) varies from 0 to \(\frac{\pi}{2}\) (covering only the upper half of the sphere), and the azimuthal angle \(\phi\) ranges from 0 to \(2\pi\), rotating around the entire sphere.
- **Radius** \(R\): Distance from the origin.- **Polar Angle** (\(\theta\)): Angle from the positive z-axis.- **Azimuthal Angle** (\(\phi\)): Angle in the xy-plane from the positive x-axis.
In the context of our given hemisphere, the radius \(R\) is 2, the polar angle \(\theta\) varies from 0 to \(\frac{\pi}{2}\) (covering only the upper half of the sphere), and the azimuthal angle \(\phi\) ranges from 0 to \(2\pi\), rotating around the entire sphere.
Parameterization
Parameterization is a process where two parameters (say \(\theta\) and \(\phi\)) are used to describe each point on a surface. It's particularly helpful when computing surface integrals.
For the hemisphere, we can use the following parameterization in spherical coordinates:
For the hemisphere, we can use the following parameterization in spherical coordinates:
- The x-coordinate is given by \(x = 2 \sin \theta \cos \phi\).
- The y-coordinate is given by \(y = 2 \sin \theta \sin \phi\).
- The z-coordinate is given by \(z = 2 \cos \theta\).
Surface Element
In spherical coordinates, a surface element \(dS\) represents an infinitesimally small piece of area on a surface. For a sphere, this is derived using the radius and angles.
The general formula for the surface element in spherical coordinates is:
The general formula for the surface element in spherical coordinates is:
- \(dS = R^2 \sin \theta \, d\theta \, d\phi\)
Hemisphere
A hemisphere is half of a sphere, cut along a plane that passes through its center. In this problem, we are focusing on the upper half, where \(z\geq 0\).
This hemisphere can be visualized as sitting above the xy-plane in three-dimensional space.
This hemisphere can be visualized as sitting above the xy-plane in three-dimensional space.
- The equation for the full sphere is \(x^2 + y^2 + z^2 = 4\).
- We only consider the part where \(z \ge 0\), hence dealing with the hemisphere.