In Exercises 9-20, use the Divergence Theorem to find the...

In Exercises $9-20,$ use the Divergence Theorem to find the outward flux of $\mathbf{F}$ across the boundary of the region $D .$ $$\mathbf{F}=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k}$$ $$\begin{array}{l}{\text { a. Cube } D :\quad \quad \quad \quad \text { The cube cut from the first octant by }} \quad\quad\quad\quad\quad\quad\quad\quad {\text { the planes } x=1, y=1, \text { and } z=1}\end{array} $$ $$b. Cube D :\quad\quad\quad\quad \quad The \quad cube \quad bounded \quad \quad \quad by \quad the \quad\quad\quad\quad\quad\quad\quad\quad\quad planes=\pm 1, y=\pm 1, \text { and } z=\pm 1$$ $$\begin{array}{c}{\text { c. Cylindrical can } D : \text { The region cut from the solid cylinder }} \quad\quad\quad\quad\quad\quad\quad\quad {x^{2}+y^{2} \leq 4 \text { by the planes } z=0 \text { and }} \\ {z=1}\end{array}$$

In Exercises $9-20,$ use the Divergence Theorem to find the outward
flux of $\mathbf{F}$ across the boundary of the region $D .$
$$\mathbf{F}=x^{2} \mathbf{i}+y^{2} \mathbf{j}+z^{2} \mathbf{k}$$
$$\begin{array}{l}{\text { a. Cube } D :\quad \quad \quad \quad \text { The cube cut from the first octant by }} \quad\quad\quad\quad\quad\quad\quad\quad {\text { the planes } x=1, y=1, \text { and } z=1}\end{array}
$$ $$b. Cube D :\quad\quad\quad\quad \quad The \quad cube \quad bounded \quad \quad \quad by \quad the \quad\quad\quad\quad\quad\quad\quad\quad\quad planes=\pm 1, y=\pm 1, \text { and } z=\pm 1$$
$$\begin{array}{c}{\text { c. Cylindrical can } D : \text { The region cut from the solid cylinder }} \quad\quad\quad\quad\quad\quad\quad\quad {x^{2}+y^{2} \leq 4 \text { by the planes } z=0 \text { and }} \\ {z=1}\end{array}$$

Key Concepts

Volume Integral

A volume integral extends the concept of integration to functions defined over three-dimensional regions. In the context of the Divergence Theorem, it is used to compute the total divergence of a vector field over a volume, thereby replacing the more complex surface flux integral with a simpler volume computation.

Vector Field

A vector field assigns a vector to every point in space, describing quantities such as velocity or force that have both magnitude and direction. Understanding the structure of a vector field is crucial for analyzing how it interacts with surfaces and volumes in various applications of calculus.

Divergence

Divergence is a differential operator that measures the rate at which a vector field spreads out (or converges) from a point. It is calculated as the sum of the partial derivatives of the field's components and indicates the presence of sources or sinks within a region.

Divergence Theorem

The Divergence Theorem, also known as Gauss's Theorem, is a fundamental result in vector calculus that relates the flux of a vector field through a closed surface to a volume integral over the region inside the surface. It allows one to convert a difficult surface integral into a more manageable volume integral of the divergence of the field.

Flux of a Vector Field

Flux represents the amount of a vector field passing through a surface. Specifically, outward flux calculates the net flow leaving a closed surface by measuring the field's normal component across the surface, and it plays a key role in applications involving conservation laws and field behavior.

Transcript

00:01 So you have the following vector field f is equal to x squared along the i direction plus y square along the j direction and c squared along the k direction so well first you want to integrate the flux the flux out of flux out of d, the cube that is described by 1 in the first octant.

00:57 So that means that x, y and z are all positive and between the planes.

01:09 X equals 1, y equals 1, and z equals 1 so how this looks is something sort of like this in the x y x y c space so it's a unit cube and line in the first thoughtant so the flux out of this cube if we call this cube c the field cube would be to integrate over the boundary, that field, that product with n, the outer normal vector to the boundary.

02:13 But it's easier to use the divergent theorem and integrate over the volume, over the inside of the cube, the divergence of f, p, bulb.

02:31 So well first what is that object what is that divergence so let's first compute what is the divergence that will be equal to differentiating with respect to xx squared plus differentiating with respect to y y squared plus differentiating with respect to c squared so that we have 2x2y plus 2 y plus 2 z so that this integral will be equal to the interval between 0 and 1 0 and 1 of x plus y plus c so for x the x the x the x the y the c so for x the x the y the c separated into sorry twice so the first integral the integral from 0 to 1 of x d x the y dc plus so 2 times all of this all of the following plus an integral of 0 to 1 0 to 1 of y d x the x the y d x the y d x the y d c plus the integral of the last function of c, the x, dx, d .c.

04:43 So first this integral, we will do that integral and that is x squared, halfs between 0 and 1 times the other intervals from 0 to 1, 0 to 1, and the y, d z, and since this don't depend on on x, those would be well just 1 times 1.

05:12 So the volume of the surface of the square, so that this would be equal to just 1 half times 1.

05:22 So that is the first one.

05:23 Now for this one we can interchange and do the integral with respect to y, which is y squared, halves, between 0 and 0.

05:33 1 times the interval between 0 and 1 0 and 1 the x the z and then in a similar way this one will be c square the integral with respect to z between 0 1 the integral between 0 and 1 the integral between 0 and 1 the x the y and so well this would be 1 half again times 1 and then this one is also 1 half times 1 so that this integral this whole integral would be equal to two times one half plus one half plus one half is equal to two times three halves that is equal to three yeah so this flux is equal to three over this first cube this flux is equal to three now now if you have the same vector field and we would like to integrate over this symmetric cube so in this situation so let's call this b.

07:08 C is described by so is the region between 4x between minus 1 and 1, y between minus 1 and 1 and 1 and c between between minus 1 and 1 so again we want to all this looks like a symmetric cube of length 2 symmetrical around the origin in every direction so you would like to compute again what is the flux of this but again we can use the that flux will be equal to derailing over the volume, for most of c the divergence of f the volume.

08:18 So well this would be the integral, so three integrals between of the function you already computed what is that divergence is equal to that so it would be equal to two times x plus y plus c, uh, z, that is a z, dx, d y, d z.

08:53 And again, we can separate it into the integrals...

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