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Calculus 3 : Line Integrals

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Example Questions

Example Question #41 : Divergence

Find the divergence of the following vector field:

$\vec{F}=\left \langle xy,yz, \cos(z)\right \rangle$ $\displaystyle \vec{F}=\left \langle xy,yz, \cos(z)\right \rangle$

Possible Answers:

$y+z-\sin(z)$ $\displaystyle y+z-\sin(z)$

$y-\sin(z)$ $\displaystyle y-\sin(z)$

$z-\sin(z)$ $\displaystyle z-\sin(z)$

$y+z+\sin(z)$ $\displaystyle y+z+\sin(z)$

Correct answer:

$y+z-\sin(z)$ $\displaystyle y+z-\sin(z)$

Explanation:

The divergence of a vector field is given by

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

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

In taking the dot product, we end up with the sum of the respective partial derivatives of the vector field. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=y $\displaystyle f_x=y$, f_y=z $\displaystyle f_y=z$. $f_z=-\sin(z)$ $\displaystyle f_z=-\sin(z)$

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Example Question #42 : Divergence

Find the divergence of the vector $\left \langle 2x^2,5y,z^3\right \rangle$ $\displaystyle \left \langle 2x^2,5y,z^3\right \rangle$

Possible Answers:

$\displaystyle 4x+5+3z^2$

$\displaystyle 4x^3+5+3z^2$

$\displaystyle 4x+5+z^2$

$\displaystyle 4x-5+3z^2$

Correct answer:

$\displaystyle 4x+5+3z^2$

Explanation:

To find the divergence of the vector $v=\left \langle P,Q,R\right \rangle$ $\displaystyle v=\left \langle P,Q,R\right \rangle$, we use the following formula

$\bigtriangledown v=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$ $\displaystyle \bigtriangledown v=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}$

Applying to the vector from the problem statement, we get

$\frac{\partial }{\partial x}(2x^2)+\frac{\partial }{\partial y}(5y)+\frac{\partial }{\partial z}(z^3)=4x+5+3z^2$ $\displaystyle \frac{\partial }{\partial x}(2x^2)+\frac{\partial }{\partial y}(5y)+\frac{\partial }{\partial z}(z^3)=4x+5+3z^2$

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Example Question #51 : Line Integrals

Find the divergence of the vector $\left \langle 3xyz,5y^2z,z^4\right \rangle$ $\displaystyle \left \langle 3xyz,5y^2z,z^4\right \rangle$

Possible Answers:

$\displaystyle 13yz+2z^3$

$\displaystyle 10yz+4z^3$

$\displaystyle 13yz+4z^3$

$\displaystyle 13z+4z^3$

Correct answer:

$\displaystyle 13yz+4z^3$

Explanation:

To find the divergence of the vector $v=\left \langle P,Q,R\right \rangle$ $\displaystyle v=\left \langle P,Q,R\right \rangle$, we use the following formula

Applying to the vector from the problem statement, we get

$\frac{\partial }{\partial x}(3xyz)+\frac{\partial }{\partial y}(5y^2z)+\frac{\partial }{\partial z}(z^4)=13yz+4z^3$ $\displaystyle \frac{\partial }{\partial x}(3xyz)+\frac{\partial }{\partial y}(5y^2z)+\frac{\partial }{\partial z}(z^4)=13yz+4z^3$

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Example Question #52 : Line Integrals

Find the divergence of the following vector field:

$\vec{F}=\left \langle z\tan(y), y, x\right \rangle$ $\displaystyle \vec{F}=\left \langle z\tan(y), y, x\right \rangle$

Possible Answers:

$\tan(y)+1$ $\displaystyle \tan(y)+1$

$\displaystyle 1$

$\displaystyle 0$

$\left \langle 0, 1, 0\right \rangle$ $\displaystyle \left \langle 0, 1, 0\right \rangle$

Correct answer:

$\displaystyle 1$

Explanation:

The divergence of a vector field is given by

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

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

When we take the dot product, we end up with the sum of the respective partial derivatives of the vector field.

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

f_x=0 $\displaystyle f_x=0$, f_y=1 $\displaystyle f_y=1$, f_z=0 $\displaystyle f_z=0$

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Example Question #43 : Divergence

Find the divergence of the following vector field:

$\vec{F}=\left \langle 5\tan(x^2), z, xy^2\right \rangle$ $\displaystyle \vec{F}=\left \langle 5\tan(x^2), z, xy^2\right \rangle$

Possible Answers:

$5x\sec^2(x^2)$ $\displaystyle 5x\sec^2(x^2)$

$10x\sec(x^2)$ $\displaystyle 10x\sec(x^2)$

$10x\sec^2(x^2)$ $\displaystyle 10x\sec^2(x^2)$

$\displaystyle 0$

Correct answer:

$10x\sec^2(x^2)$ $\displaystyle 10x\sec^2(x^2)$

Explanation:

The divergence of a vector field is given by

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

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

When we take the dot product, we end up with the sum of the respective partial derivatives of the vector field.

To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=10x\sec^2(x^2)$ $\displaystyle f_x=10x\sec^2(x^2)$, f_y=0 $\displaystyle f_y=0$, f_z=0 $\displaystyle f_z=0$

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Example Question #1701 : Calculus 3

Find the divergence of the following vector field:

$\vec{F}=\left \langle 7e^{3x}, 2yz, y^3\right \rangle$ $\displaystyle \vec{F}=\left \langle 7e^{3x}, 2yz, y^3\right \rangle$

Possible Answers:

$21e^{3x}+2z$ $\displaystyle 21e^{3x}+2z$

$21e^{3x}+2z+3y^2$ $\displaystyle 21e^{3x}+2z+3y^2$

$21e^{3x}+2z+y^3$ $\displaystyle 21e^{3x}+2z+y^3$

$21e^{x}+2z$ $\displaystyle 21e^{x}+2z$

Correct answer:

$21e^{3x}+2z$ $\displaystyle 21e^{3x}+2z$

Explanation:

The divergence of the vector field is given by

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

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

Taking the dot product gives us the sum of the respective partial derivatives of the vector field. For higher order partial derivatives, we work from left to right for the given variables.

The partial derivatives are

$f_x=21e^{3x}$ $\displaystyle f_x=21e^{3x}$, f_y=2z $\displaystyle f_y=2z$, f_z=0 $\displaystyle f_z=0$

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Example Question #51 : Divergence

Find the divergence of the following vector field:

$\vec{F}=\left \langle e^{\tan(xz) }, e^{y^2}, z^3\right \rangle$ $\displaystyle \vec{F}=\left \langle e^{\tan(xz) }, e^{y^2}, z^3\right \rangle$

Possible Answers:

$x\sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$ $\displaystyle x\sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$

$\sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$ $\displaystyle \sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$

$z\sec(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$ $\displaystyle z\sec(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$

$z\sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$ $\displaystyle z\sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$

$-z\sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$ $\displaystyle -z\sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$

Correct answer:

$z\sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$ $\displaystyle z\sec^2(xz)e^{\tan(xz)}+2ye^{y^2}+3z^2$

Explanation:

The divergence of the vector field is given by

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

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

Taking the dot product gives us the sum of the respective partial derivatives of the vector field. For higher order partial derivatives, we work from left to right for the given variables.

The partial derivatives are

$f_x=z\sec^2(xz)e^{\tan(xz)}$ $\displaystyle f_x=z\sec^2(xz)e^{\tan(xz)}$, $f_y=2ye^{y^2}$ $\displaystyle f_y=2ye^{y^2}$, $\displaystyle f_z=3z^2$

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Example Question #52 : Line Integrals

Find the divergence of the vector $\left \langle 3x,2xy^2,z^4\right \rangle$ $\displaystyle \left \langle 3x,2xy^2,z^4\right \rangle$

Possible Answers:

$\displaystyle 3+4xy+4z^3$

$\displaystyle 5+xy+4z^3$

$\displaystyle 7xy+4z^3$

3+4xy $\displaystyle 3+4xy$

Correct answer:

$\displaystyle 3+4xy+4z^3$

Explanation:

To find the divergence of a vector $v=\left \langle P,Q,R\right \rangle$ $\displaystyle v=\left \langle P,Q,R\right \rangle$, we apply the formula:

$div(v)=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R }{\partial z}$ $\displaystyle div(v)=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R }{\partial z}$

Using the vector from the problem statement, we get

$\frac{\partial }{\partial x}(3x)+\frac{\partial }{\partial y}(2xy^2)+\frac{\partial }{\partial z}(z^4)=3+4xy+4z^3$ $\displaystyle \frac{\partial }{\partial x}(3x)+\frac{\partial }{\partial y}(2xy^2)+\frac{\partial }{\partial z}(z^4)=3+4xy+4z^3$

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Example Question #51 : Divergence

Find the divergence of the vector $\left \langle \sin(x),x^4y,xyz\right \rangle$ $\displaystyle \left \langle \sin(x),x^4y,xyz\right \rangle$

Possible Answers:

$\cos(x)+4x^3y+y$ $\displaystyle \cos(x)+4x^3y+y$

$-\cos(x)+4x^3y+xy$ $\displaystyle -\cos(x)+4x^3y+xy$

$\cos(x)+x^4+xy$ $\displaystyle \cos(x)+x^4+xy$

$\cos(x)+x^3y+xy$ $\displaystyle \cos(x)+x^3y+xy$

Correct answer:

$\cos(x)+x^4+xy$ $\displaystyle \cos(x)+x^4+xy$

Explanation:

To find the divergence of a vector $v=\left \langle P,Q,R\right \rangle$ $\displaystyle v=\left \langle P,Q,R\right \rangle$, we apply the formula:

Using the vector from the problem statement, we get

$\frac{\partial }{\partial x}(\sin(x))+\frac{\partial }{\partial y}(x^4y)+\frac{\partial }{\partial z}(xyz)=\cos(x)+x^4+xy$ $\displaystyle \frac{\partial }{\partial x}(\sin(x))+\frac{\partial }{\partial y}(x^4y)+\frac{\partial }{\partial z}(xyz)=\cos(x)+x^4+xy$

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Example Question #52 : Divergence

Find the divergence of the following vector field:

$\vec{F}=\left \langle y^3z, xy^2, e^{xyz}\right \rangle$ $\displaystyle \vec{F}=\left \langle y^3z, xy^2, e^{xyz}\right \rangle$

Possible Answers:

$2xy+xye^{xyz}$ $\displaystyle 2xy+xye^{xyz}$

$2xy+e^{xyz}$ $\displaystyle 2xy+e^{xyz}$

$2xy+yze^{xyz}$ $\displaystyle 2xy+yze^{xyz}$

$y^3+2xy+xye^{xyz}$ $\displaystyle y^3+2xy+xye^{xyz}$

Correct answer:

$2xy+xye^{xyz}$ $\displaystyle 2xy+xye^{xyz}$

Explanation:

The divergence of the vector field

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

where $\bigtriangledown = \left \langle \frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z}\right \rangle$ $\displaystyle \bigtriangledown = \left \langle \frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z}\right \rangle$

In taking the dot product, we get the sum of the respective partial derivatives of the vector field.

The partial derivatives are

f_x=0 $\displaystyle f_x=0$, f_y=2xy $\displaystyle f_y=2xy$, $f_z=xye^{xyz}$ $\displaystyle f_z=xye^{xyz}$

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Calculus 3 : Line Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #41 : Divergence

Example Question #42 : Divergence

Example Question #51 : Line Integrals

Example Question #52 : Line Integrals

Example Question #43 : Divergence

Example Question #1701 : Calculus 3

Example Question #51 : Divergence

Example Question #52 : Line Integrals

Example Question #51 : Divergence

Example Question #52 : Divergence

All Calculus 3 Resources

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