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Calculus 3 : Line Integrals

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Example Questions

Example Question #21 : Divergence

Find the divergence of the vector $\left \langle \sin(x),\tan(y),2yz^2\right \rangle$

Possible Answers:

$\cos(x)-\sec^2(y)+4yz$

$\sin(x)+\sec^2(y)+4yz$

$-\cos(x)+\sec(y)+4yz$

$\cos(x)+\sec^2(y)+4yz$

Correct answer:

$\cos(x)+\sec^2(y)+4yz$

Explanation:

To find the divergence of a vector $v=\left \langle A,B,C\right \rangle$ , we apply the following definition: $div(v)=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$ . Applying the definition to the vector from the problem statement, we get $\frac{\partial }{\partial x}(\sin(x))+\frac{\partial }{\partial y}(\tan(x))+\frac{\partial }{\partial z}(2yz^2)=\cos(x)+\sec^2(y)+4yz$

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Example Question #31 : Line Integrals

Find $div \vec{F}$ , where F is given by

$\left \langle x^2z, z^2, ye^xz\right \rangle$

Possible Answers:

ye^x

2xz+ye^x

2xz

Correct answer:

2xz+ye^x

Explanation:

The divergence of a vector is given by

$\nabla \cdot \vec{F}$ , where $\nabla = \left \langle f_x, f_y, f_z\right \rangle$

Taking the partial respective partial derivatives of the x, y, and z components of our curve, we get

f_x=2xz

f_y=0

f_z=ye^x

The rules used to find the derivatives are as follows:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #1681 : Calculus 3

Calculate the divergence of the following vector:

$\mathbf{F}=x^2y\mathbf{i}+3xz\mathbf{j}+z^3\mathbf{k}$

Possible Answers:

2xy+3z^2

x^2+3

2xy+3+3z

2x^2y+3x

x^2+3z^2

Correct answer:

2xy+3z^2

Explanation:

For a given vector

$\mathbf{F}=U\mathbf{i}+V\mathbf{k}+W\mathbf{k}$

The divergence is calculated by:

$\triangledown\cdot \mathbf{F}=\frac{\partial U}{\partial x} +\frac{\partial V}{\partial y}+\frac{\partial W}{\partial z}$

For our vector

$\triangledown\cdot \mathbf{F}=2xy+0+3z^2$

The answer is, therefore:

2xy+3z^2

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Example Question #1681 : Calculus 3

Given that F is a vector function and f is a scalar function, which of the following operations results in a vector?

Possible Answers:

$\nabla \cdot(\nabla f)$

$\nabla(\nabla \cdot f)$

$\nabla \cdot (\nabla\cdot \mathbf{F})$

$\nabla \cdot (\nabla \times \mathbf{F})$

$\nabla \times (\nabla \times \mathbf{F})$

Correct answer:

$\nabla \times (\nabla \times \mathbf{F})$

Explanation:

For all the given answers:

$\nabla \cdot (\nabla \times \mathbf{F})$ - The curl of a vector is also a vector, so the divergence of the term in parenthesis is scalar.

$\nabla \cdot (\nabla \cdot \mathbf{F})$ - The divergence of a vector is a scalar. The divergence of a scalar is undefined, so this expression is undefined.

$\nabla \cdot(\nabla f)$ - The gradient of a scalar is a vector, so the divergence of the term in parenthesis is a scalar.

$\nabla(\nabla \cdot f)$ - The divergence of a scalar doesn't exist, so this expression is undefined.

The remaining answer is:

$\nabla \times (\nabla \times \mathbf{F})$ - The curl of a vector is also a vector, so the curl of the term in parenthesis is a vector as well.

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Example Question #1682 : Calculus 3

Given that F is a vector function and f is a scalar function, which of the following expressions is valid?

Possible Answers:

$\nabla\cdot(\nabla\cdot\mathbf{F})$

$\nabla(\nabla \cdot f)$

$\nabla\cdot(\nabla\mathbf{F})$

$\nabla \cdot (\nabla \times f)$

$\nabla \cdot (\nabla f)$

Correct answer:

$\nabla \cdot (\nabla f)$

Explanation:

For each of the given answers:

$\nabla \cdot (\nabla \times f)$ - The curl of a scalar is undefined, so the term in parenthesis is invalid.

$\nabla(\nabla \cdot f)$ - The divergence of a scalar is also undefined, so the term in parenthesis is invalid.

$\nabla\cdot(\nabla\mathbf{F})$ - The gradient of a scalar is undefined as well, so the term in parenthesis is invalid.

$\nabla\cdot(\nabla\cdot\mathbf{F})$ - The divergence of a vector is a scalar. The divergence of a the term in parenthesis, which is a scalar, is undefined, so the expression is invalid.

The remaining answer must be correct:

$\nabla \cdot (\nabla f)$ - The gradient of a scalar is a vector, so the divergence of the term in parenthesis is a scalar. The expression is valid.

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Example Question #26 : Divergence

Compute the divergence of the following vector function:

$\mathbf{F}=3yz\mathbf{i}-xy\mathbf{j}+(3xz+y^3)\mathbf{k}$

Possible Answers:

-y+3z+3y^2

3z+4x

3z-x

Correct answer:

Explanation:

For a vector function $\mathbf{F}=U\mathbf{i}+V\mathbf{j}+W\mathbf{k}$ ,

the divergence is defined by:

$\nabla \cdot \mathbf{F}=\frac{\partial U}{\partial x}+\frac{\partial V}{\partial y} +\frac{\partial W}{\partial z}$

For our function:

$\frac{\partial U}{\partial x}=\frac{\partial }{\partial x}(3yz)=0$

$\frac{\partial V}{\partial y}=\frac{\partial }{\partial y}(-xy)=-x$

$\frac{\partial W}{\partial z}=\frac{\partial }{\partial z}(3xz+y^3)=3x$

Thus, the divergence of our function is:

$\nabla \cdot \mathbf{F}=-0-x+3x=2x$

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Example Question #31 : Divergence

Compute the divergence of the following vector function:

$\mathbf{F}=\cos(xy)\mathbf{i}+4y^3z^2\mathbf{j}+5xe^{2z}\mathbf{k}$

Possible Answers:

$-y\sin(xy)+12y^2z^2+10xe^{2z}$

$-\sin(xy)+12y^2z^2+5xe^{2z}$

$y\sin(x)+12y^2z^3+10xe^{z}$

$-x\cos(xy)+16y^2z^2+5e^{2z}$

Correct answer:

$-y\sin(xy)+12y^2z^2+10xe^{2z}$

Explanation:

For a vector function $\mathbf{F}=U\mathbf{i}+V\mathbf{j}+W\mathbf{k}$ ,

the divergence is defined by:

$\nabla \cdot \mathbf{F}=\frac{\partial U}{\partial x}+\frac{\partial V}{\partial y} +\frac{\partial W}{\partial z}$

For our function:

$\frac{\partial U}{\partial x}=\frac{\partial }{\partial x}(\cos(xy))=-y\sin(xy)$

$\frac{\partial V}{\partial y}=\frac{\partial }{\partial y}(4y^3z^2)=12y^2z^2$

$\frac{\partial W}{\partial z}=\frac{\partial }{\partial z}(5xe^{2z})=10xe^{2z}$

Thus, the divergence of our function is:

$\nabla \cdot \mathbf{F}=-y\sin(xy)+12y^2z^2+10xe^{2z}$

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Example Question #32 : Line Integrals

Find $div \vec{F}$ , where F is given by the following:

$\vec{F}=\left \langle y\cos(z), \frac{y}{x}, z^3\cos(z)\right \rangle$

Possible Answers:

$\frac{1}{x}+3z^2\cos(z)-z^3\sin(z)$

$\left \langle 0, \frac{1}{x}, 3z^2\cos(z)-z^3\sin(z)\right \rangle$

$\frac{1}{x}+3z^2\cos(z)+z^3\sin(z)$

Correct answer:

$\frac{1}{x}+3z^2\cos(z)-z^3\sin(z)$

Explanation:

The divergence of a vector function is given by

$div \vec{F}= \bigtriangledown \cdot \vec{F}$

where $\bigtriangledown = \left \langle f_x, f_y, f_z\right \rangle$

So, we must find the partial derivative of each respective component.

The partial derivatives are

f_x=0

$f_y=\frac{1}{x}$

$f_z=3z^2\cos(z)-z^3\sin(z)$

The partial derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #33 : Divergence

Find $div\vec{F}$ , where F is given by the following curve:

$\vec{F}=\left \langle y^2, y^3, z^2 \cos(xy)\right \rangle$

Possible Answers:

$2y+3y^2-xz^2\sin(xy)$

$3y^2+2z\cos(xy)$

$\left \langle 0, 3y^2, 2z\cos(xy)\right \rangle$

Correct answer:

$3y^2+2z\cos(xy)$

Explanation:

The divergence of a curve is given by

$div\vec{F}=\nabla \cdot \vec{F}$

where $\nabla = \left \langle f_x, f_y, f_z\right \rangle$

So, we must find the partial derivatives of the x, y, and z components, respectively:

f_x=0

f_y=3y^2

$f_z=2z\cos(xy)$

The partial derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #34 : Divergence

Find $div \vec{F}$ of the given function:

$\vec{F}=\left \langle 4xy, y\cos(y), xy\right \rangle$

Possible Answers:

$4y+\cos(y)+y\cos(y)$

$\cos(y)+y(4+\sin(y))$

$\cos(y)+y(4-\sin(y))$

$\cos(y)+4-\sin(y)$

Correct answer:

$\cos(y)+y(4-\sin(y))$

Explanation:

The divergence of a vector function is given by

$div \vec{F}=\bigtriangledown \cdot \vec{F}$

where $\bigtriangledown = \left \langle f_x, f_y, f_z\right \rangle$

So, we take the respective partial derivatives of the x, y, and z-components of the vector function, and add them together (from the dot product):

f_x=4y

$f_y=\cos(y)-y\sin(y)$

f_z=0

The partial derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Calculus 3 : Line Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #21 : Divergence

Example Question #31 : Line Integrals

Example Question #1681 : Calculus 3

Example Question #1681 : Calculus 3

Example Question #1682 : Calculus 3

Example Question #26 : Divergence

Example Question #31 : Divergence

Example Question #32 : Line Integrals

Example Question #33 : Divergence

Example Question #34 : Divergence

All Calculus 3 Resources

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