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Calculus 3 : Line Integrals

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Example Questions

Example Question #31 : Divergence

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

$\vec{F}=\cos(x)\hat{i}+(xyz)\hat{j}+z^2\sin(y)\hat{k}$

Possible Answers:

$-\sin(x)+xz-2z\sin(y)$

$\sin(x)+xz+2z\sin(y)$

$\left \langle -\sin(x),xz,2z\sin(y) \right \rangle$

$-\sin(x)+xz+2z\sin(y)$

Correct answer:

$-\sin(x)+xz+2z\sin(y)$

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, in taking the dot product of the gradient and the function, we get the sum of the respective partial derivatives. To find the given partial derivative of the function, we must treat the other variable(s) as constants.

The partial derivatives are

$f_x=-\sin(x)$

f_y=xz

$f_z=2z\sin(y)$

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Example Question #32 : Divergence

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

$\vec{F}=\left \langle xe^{\cos(y)+\sin(z)}, 2y, z\right \rangle$

Possible Answers:

$e^{\cos(y)+\sin(z)}$

$e^{\cos(y)+\sin(z)}+2y+1$

$e^{\cos(y)+\sin(z)}+2$

$e^{\cos(y)+\sin(z)}+3$

Correct answer:

$e^{\cos(y)+\sin(z)}+3$

Explanation:

The divergence of a vector function is given by

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

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

The partial derivatives are

$f_x=e^{\cos(y)+\sin(z)}$

f_y=2

f_z=1

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Example Question #37 : Divergence

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

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The divergence of a vector field is given by

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

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

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

The partial derivatives are

$f_x=2x\cos(xy)-x^2y\sin(xy)$

f_y=z^3

f_z=0

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Example Question #1691 : Calculus 3

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

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

Possible Answers:

$\frac{y}{x}+2xy^2z+\frac{1}{z}$

$-\frac{y}{x}+2x^2yz-\frac{1}{z}$

$\frac{y}{x}+2x^2yz+\frac{1}{z}$

$\frac{1}{x}+2x^2yz+\frac{1}{z}$

Correct answer:

$\frac{y}{x}+2x^2yz+\frac{1}{z}$

Explanation:

The divergence of a vector field is given by

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

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

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

The partial derivatives are

$f_x=\frac{y}{x}$ , f_y=2x^2yz , $f_z=\frac{1}{z}$

Report an Error

Example Question #1692 : Calculus 3

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

$\vec{F}=\left \langle \sec(5x), \ln(z), xyz\right \rangle$

Possible Answers:

$-5\sec(5x)\tan(5x)+\frac{1}{z}+xy$

$5\sec(5x)\tan(5x)+\frac{1}{z}+xy$

$5\sec(5x)\tan(5x)+\frac{1}{z}+xyz$

$\sec(5x)\tan(5x)+\frac{1}{z}+xy$

Correct answer:

$5\sec(5x)\tan(5x)+\frac{1}{z}+xy$

Explanation:

The divergence of a vector field is given by

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

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

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

The partial derivatives are

$f_x=5\sec(5x)\tan(5x)$

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

f_z=xy

Report an Error

Example Question #1693 : Calculus 3

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

Possible Answers:

$4xy^3+7y^4+\sin(z)$

$4xy^2+7y^4-\sin(z)$

$xy^3+5y^4-\sin(z)$

$4xy^3+5y^4-\sin(z)$

Correct answer:

$4xy^3+5y^4-\sin(z)$

Explanation:

To find the divergence of a vector $a=\left \langle A,B,C\right \rangle$ , we use the definition

$divergence=\frac{\partial A}{\partial x}+\frac{\partial B}{\partial y}+\frac{\partial C}{\partial z}$

Using the vector from the problem statement, we get

$divergence=\frac{\partial }{\partial x}(2x^2y^3)+\frac{\partial }{\partial y}(y^5)+\frac{\partial }{\partial z}(\cos(z))=4xy^3+5y^4-\sin(z)$

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Example Question #1694 : Calculus 3

Find $div \vec{F}$ of the vector field:

$\vec{F}=\left \langle 3z, 4x, 5y\right \rangle$

Possible Answers:

3z+4x+5y

Correct answer:

Explanation:

The divergence of a vector field is given by

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

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

In taking the dot product, we are left with the sum of the respective partial derivatives of the vector function. 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 , f_y=0 , f_z=0

Report an Error

Example Question #1695 : Calculus 3

Find $div \vec{F}$ of the vector field:

$\vec{F}=\left \langle 3xy, 5y\sin(z), 2\right \rangle$

Possible Answers:

$3y+5\sin(z)+2$

$3y+5\sin(z)$

3x+5y

$3y-5\sin(z)$

Correct answer:

$3y+5\sin(z)$

Explanation:

The divergence of a vector field is given by

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

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

The partial derivatives are

f_x=3y , $f_y=5\sin(z)$ , f_z=0

Report an Error

Example Question #41 : Line Integrals

Find $div \vec{F}$ , where F is the following vector field:

$\vec{F}=\left \langle e^{xyz}, xy, \sec(z)\right \rangle$

Possible Answers:

$yze^{xyz}+x+\sec(z)\tan(z)$

$yze^{xyz}+x-\sec(z)\tan(z)$

$yze^{xyz}+xy+\sec(z)\tan(z)$

Correct answer:

$yze^{xyz}+x+\sec(z)\tan(z)$

Explanation:

The divergence of a vector field is given by

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

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

In taking the dot product, we get 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

$f_x=yze^{xyz}$

f_y=x

$f_z=\sec(z)\tan(z)$

Report an Error

Example Question #42 : Line Integrals

Find $div \vec{F}$ , where F is the following vector field:

$\vec{F}=\left \langle x\tan(z^2), 3yz, 8e^{xz}\right \rangle$

Possible Answers:

$\tan(z^2)+3z+8xe^{xz}$

$\tan(z^2)+3y+8xe^{xz}$

$\sec^2(z^2)+3z+8xe^{xz}$

$\tan(z^2)+3z+8e^{xz}$

Correct answer:

$\tan(z^2)+3z+8xe^{xz}$

Explanation:

The divergence of a vector field is given by

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

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

The partial derivatives are

$f_x=\tan(z^2)$

f_y=3z

$f_z=8xe^{xz}$

Report an Error

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Calculus 3 : Line Integrals

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #31 : Divergence

Example Question #32 : Divergence

Example Question #37 : Divergence

Example Question #1691 : Calculus 3

Example Question #1692 : Calculus 3

Example Question #1693 : Calculus 3

Example Question #1694 : Calculus 3

Example Question #1695 : Calculus 3

Example Question #41 : Line Integrals

Example Question #42 : Line Integrals

All Calculus 3 Resources

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