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Magazine Chapter 8: Problem 16
Evaluate the surface integral \(\iint_{\partial S} \mathbf{F} \cdot \mathbf{n} d A,\) where \(\mathbf{F}(x, y, z)=\mathbf{i}+\mathbf{j}+z\left(x^{2}+y^{2}\right)^{2} \mathbf{k}\) and \(\partial S\) is the surface of the cylinder \(x^{2}+y^{2} \leq 1,0 \leq z \leq 1.\)
Short Answer
Expert verified
The value of the surface integral is \(\frac{\pi}{3}\).
Step by step solution
01
Understand the Problem
First, identify the components of the vector field \( \mathbf{F}(x, y, z) = \mathbf{i} + \mathbf{j} + z(x^2 + y^2)^2 \mathbf{k} \) and the surface \( \partial S \) which is the curved surface and two circular ends of the cylinder \( x^2+y^2 \leq 1, 0 \leq z \leq 1 \).
02
Apply the Divergence Theorem
To simplify the evaluation of the surface integral, use the Divergence Theorem, which states: \[ \iint_{\partial S} \mathbf{F} \cdot \mathbf{n} \, dA = \iiint_S abla \cdot \mathbf{F} \, dV, \] where \(S\) is the volume bounded by \(\partial S\), and \(abla \cdot \mathbf{F}\) is the divergence of \(\mathbf{F}\).
03
Calculate the Divergence \(\nabla \cdot \mathbf{F}\)
Compute the divergence of \(\mathbf{F}\): \[ abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(1) + \frac{\partial}{\partial y}(1) + \frac{\partial}{\partial z}\left[z(x^2 + y^2)^2\right]. \] Noting that \(\frac{\partial}{\partial x}(1) = 0\) and \(\frac{\partial}{\partial y}(1) = 0\), compute \( \frac{\partial}{\partial z}(z(x^2 + y^2)^2) = (x^2 + y^2)^2 \). Hence, \( abla \cdot \mathbf{F} = (x^2 + y^2)^2. \)
04
Set Up and Evaluate the Triple Integral
The volume integral becomes \[ \iiint_S (x^2 + y^2)^2 \, dV = \int_0^1 \int_{x^2+y^2 \leq 1} (x^2 + y^2)^2 \, dx \, dy \, dz. \] Convert to cylindrical coordinates: \( x = r\cos\theta, y = r\sin\theta, dA = r \, dr \, d\theta \). Thus, the integral becomes \[ \int_0^1 \int_0^{2\pi} \int_0^1 r^5 \, dr \, d\theta \, dz. \]
05
Compute the Integrals
First, compute the integral with respect to \( r \): \[ \int_0^1 r^5 \, dr = \frac{r^6}{6} \bigg|_0^1 = \frac{1}{6}. \] Next, compute the integral with respect to \( \theta \): \[ \int_0^{2\pi} \, d\theta = 2\pi. \] Finally, compute the integral with respect to \( z \): \[ \int_0^1 \, dz = 1. \] Thus, the entire integral evaluates to \( \frac{1}{6} \times 2\pi \times 1 = \frac{\pi}{3}. \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Surface Integral
A surface integral allows us to extend the concept of a line integral to higher dimensions, letting us integrate over a surface instead of a path. In this case, we're interested in the surface integral of a vector field over a specific surface.
The notation \( \iint_{\partial S} \mathbf{F} \cdot \mathbf{n} \, dA \) represents the surface integral of a vector field \( \mathbf{F} \) over the surface \( \partial S \). Here, \( \mathbf{n} \) is the unit normal vector to the surface, and \( dA \) is an infinitesimal area element on \( \partial S \). Essentially, this integral computes the "flow" of the vector field through a surface.
The notation \( \iint_{\partial S} \mathbf{F} \cdot \mathbf{n} \, dA \) represents the surface integral of a vector field \( \mathbf{F} \) over the surface \( \partial S \). Here, \( \mathbf{n} \) is the unit normal vector to the surface, and \( dA \) is an infinitesimal area element on \( \partial S \). Essentially, this integral computes the "flow" of the vector field through a surface.
- The vector field \( \mathbf{F}(x, y, z) = \mathbf{i} + \mathbf{j} + z(x^2 + y^2)^2 \mathbf{k} \) is defined over the surface, which is a cylinder including its top and bottom surfaces.
- The challenge lies in considering all the segments of the surface to account for the entire flow through the cylinder's boundary.
Divergence Theorem
The Divergence Theorem is a powerful tool in vector calculus, linking surface integrals over a closed surface to volume integrals over the region inside.
It states: \( \iint_{\partial S} \mathbf{F} \cdot \mathbf{n} \, dA = \iiint_S abla \cdot \mathbf{F} \, dV \), meaning the outward flux of a vector field through a closed surface equals the volume integral of the divergence over the region it encloses.
It states: \( \iint_{\partial S} \mathbf{F} \cdot \mathbf{n} \, dA = \iiint_S abla \cdot \mathbf{F} \, dV \), meaning the outward flux of a vector field through a closed surface equals the volume integral of the divergence over the region it encloses.
- This theorem simplifies the computation significantly, as evaluating a triple integral often turns out to be more straightforward than dealing with multiple surface integrals.
- In this exercise, the surface \( \partial S \) is a cylinder with a volume \( S \) inside it, bounded by the specified limits.
- By utilizing the divergence theorem, we switch from having to deal with the surface integral directly and focus on computing a triple integral after finding the divergence, \( abla \cdot \mathbf{F} \).
Cylindrical Coordinates
Cylindrical coordinates simplify integrals over symmetrical shapes like cylinders by representing points in a three-dimensional space using radius \( r \), angle \( \theta \), and height \( z \).
Common conversions include \( x = r\cos\theta \) and \( y = r\sin\theta \). Volume and surface elements adapt to \( dA = r \, dr \, d\theta \) for surface integrals, and \( dV = r \, dr \, d\theta \, dz \) for volume integrals.
Common conversions include \( x = r\cos\theta \) and \( y = r\sin\theta \). Volume and surface elements adapt to \( dA = r \, dr \, d\theta \) for surface integrals, and \( dV = r \, dr \, d\theta \, dz \) for volume integrals.
- These coordinates are especially relevant to our problem, because the geometry of the surface \( \partial S \) (a cylinder) can be naturally expressed in these terms.
- The integration limits are also straightforward: \( r \) goes from 0 to 1, \( \theta \) from 0 to \( 2\pi \), and \( z \) from 0 to 1.
- Using cylindrical coordinates reduces the complexity of the expression \((x^2 + y^2)^2\) to \( r^4 \), further simplifying our calculations.
Triple Integral
A triple integral extends the concept of integrating functions over a three-dimensional region or volume. For our exercise, this means evaluating the integral over the cylindrical volume we transformed using cylindrical coordinates.
The triple integral \( \iiint_S (x^2 + y^2)^2 \, dV \) becomes \[ \int_0^1 \int_0^{2\pi} \int_0^1 r^5 \, dr \, d\theta \, dz. \]
The triple integral \( \iiint_S (x^2 + y^2)^2 \, dV \) becomes \[ \int_0^1 \int_0^{2\pi} \int_0^1 r^5 \, dr \, d\theta \, dz. \]
- The integration over \( r \) results in \( \frac{1}{6} \), by evaluating \( \int_0^1 r^5 \, dr \).
- Integrating over \( \theta \) gives \( 2\pi \), by evaluating \( \int_0^{2\pi} \, d\theta \).
- The integration over \( z \) is straightforward and results in 1, which simplifies the process significantly.
- All in all, the computed triple integral evaluates to \( \frac{\pi}{3} \), which confirms the result derived using the divergence theorem.
