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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #353 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (11,24,15) . What is the position of this point in spherical coordinates?

Possible Answers:

$(26.4,114.62^{\circ},-119.6^{\circ})$

$(30.36,114.62^{\circ},60.4^{\circ})$

$(26.4,65.38^{\circ},60.4^{\circ})$

$(30.36,65.38^{\circ},60.4^{\circ})$

$(26.4,114.62^{\circ},119.6^{\circ})$

Correct answer:

$(30.36,65.38^{\circ},60.4^{\circ})$

Explanation:

When given Cartesian coordinates of the form (x,y,z) to cylindrical coordinates of the form $(\rho,\theta,\phi)$ , it would be useful to calculate the $\rho$ term first, as we'll derive $\phi$ from it.

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

It is something to bear in mind when making a calculation using a calculator; negative values by convention create a negative $\theta$ , while negative values lead to $|\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(11,24,15)\\&\rho=\sqrt{(11)^2+(24)^2+(15)^2}=30.36\\&\theta=arctan(\frac{24}{11})=65.38^{\circ}\\&\phi=arccos(\frac{15}{30.36})=60.4^{\circ}\end{align*}$

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Example Question #354 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (-8,-11,15) . What is the position of this point in spherical coordinates?

Possible Answers:

$(13.6,53.97^{\circ},137.8^{\circ})$

$(20.25,53.97^{\circ},42.2^{\circ})$

$(13.6,-126.03^{\circ},42.2^{\circ})$

$(13.6,53.97^{\circ},-137.8^{\circ})$

$(20.25,-126.03^{\circ},42.2^{\circ})$

Correct answer:

$(20.25,-126.03^{\circ},42.2^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(-8,-11,15)\\&\rho=\sqrt{(-8)^2+(-11)^2+(15)^2}=20.25\\&\theta=arctan(\frac{-11}{-8})=-126.03^{\circ}\\&\phi=arccos(\frac{15}{20.25})=42.2^{\circ}\end{align*}$

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Example Question #355 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (-26,-8,2) . What is the position of this point in spherical coordinates?

Possible Answers:

$(27.2,17.1^{\circ},94.2^{\circ})$

$(27.28,-162.9^{\circ},85.8^{\circ})$

$(27.2,-162.9^{\circ},85.8^{\circ})$

$(27.28,17.1^{\circ},85.8^{\circ})$

$(27.2,17.1^{\circ},-94.2^{\circ})$

Correct answer:

$(27.28,-162.9^{\circ},85.8^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(-26,-8,2)\\&\rho=\sqrt{(-26)^2+(-8)^2+(2)^2}=27.28\\&\theta=arctan(\frac{-8}{-26})=-162.9^{\circ}\\&\phi=arccos(\frac{2}{27.28})=85.8^{\circ}\end{align*}$

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Example Question #356 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (10,-48,-42) . What is the position of this point in spherical coordinates?

Possible Answers:

$(49.03,-78.23^{\circ},130.58^{\circ})$

$(49.03,101.77^{\circ},49.42^{\circ})$

$(64.56,101.77^{\circ},130.58^{\circ})$

$(49.03,101.77^{\circ},-49.42^{\circ})$

$(64.56,-78.23^{\circ},130.58^{\circ})$

Correct answer:

$(64.56,-78.23^{\circ},130.58^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(10,-48,-42)\\&\rho=\sqrt{(10)^2+(-48)^2+(-42)^2}=64.56\\&\theta=arctan(\frac{-48}{10})=-78.23^{\circ}\\&\phi=arccos(\frac{-42}{64.56})=130.58^{\circ}\end{align*}$

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Example Question #357 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (8,41,-38) . What is the position of this point in spherical coordinates?

Possible Answers:

$(56.47,101.04^{\circ},132.29^{\circ})$

$(41.77,101.04^{\circ},-47.71^{\circ})$

$(41.77,101.04^{\circ},47.71^{\circ})$

$(56.47,78.96^{\circ},132.29^{\circ})$

$(41.77,78.96^{\circ},132.29^{\circ})$

Correct answer:

$(56.47,78.96^{\circ},132.29^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(8,41,-38)\\&\rho=\sqrt{(8)^2+(41)^2+(-38)^2}=56.47\\&\theta=arctan(\frac{41}{8})=78.96^{\circ}\\&\phi=arccos(\frac{-38}{56.47})=132.29^{\circ}\end{align*}$

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Example Question #358 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (-3,2,-28) . What is the position of this point in spherical coordinates?

Possible Answers:

$(28.23,33.69^{\circ},172.66^{\circ})$

$(28.23,146.31^{\circ},172.66^{\circ})$

$(3.61,33.69^{\circ},-7.34^{\circ})$

$(3.61,33.69^{\circ},7.34^{\circ})$

$(3.61,146.31^{\circ},172.66^{\circ})$

Correct answer:

$(28.23,146.31^{\circ},172.66^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(-3,2,-28)\\&\rho=\sqrt{(-3)^2+(2)^2+(-28)^2}=28.23\\&\theta=arctan(\frac{2}{-3})=146.31^{\circ}\\&\phi=arccos(\frac{-28}{28.23})=172.66^{\circ}\end{align*}$

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Example Question #11 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-44,-37,-7) . What is the position of this point in spherical coordinates?

Possible Answers:

$(57.49,40.06^{\circ},-83.06^{\circ})$

$(57.91,40.06^{\circ},96.94^{\circ})$

$(57.91,-139.94^{\circ},96.94^{\circ})$

$(57.49,40.06^{\circ},83.06^{\circ})$

$(57.49,-139.94^{\circ},96.94^{\circ})$

Correct answer:

$(57.91,-139.94^{\circ},96.94^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(-44,-37,-7)\\&\rho=\sqrt{(-44)^2+(-37)^2+(-7)^2}=57.91\\&\theta=arctan(\frac{-37}{-44})=-139.94^{\circ}\\&\phi=arccos(\frac{-7}{57.91})=96.94^{\circ}\end{align*}$

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Example Question #12 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-24,-34,24) . What is the position of this point in spherical coordinates?

Possible Answers:

$(41.62,54.78^{\circ},-119.97^{\circ})$

$(48.04,-125.22^{\circ},60.03^{\circ})$

$(48.04,54.78^{\circ},60.03^{\circ})$

$(41.62,-125.22^{\circ},60.03^{\circ})$

$(41.62,54.78^{\circ},119.97^{\circ})$

Correct answer:

$(48.04,-125.22^{\circ},60.03^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(-24,-34,24)\\&\rho=\sqrt{(-24)^2+(-34)^2+(24)^2}=48.04\\&\theta=arctan(\frac{-34}{-24})=-125.22^{\circ}\\&\phi=arccos(\frac{24}{48.04})=60.03^{\circ}\end{align*}$

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Example Question #13 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (34,43,-21) . What is the position of this point in spherical coordinates?

Possible Answers:

$(58.7,51.67^{\circ},110.96^{\circ})$

$(54.82,128.33^{\circ},69.04^{\circ})$

$(58.7,128.33^{\circ},110.96^{\circ})$

$(54.82,51.67^{\circ},110.96^{\circ})$

$(54.82,128.33^{\circ},-69.04^{\circ})$

Correct answer:

$(58.7,51.67^{\circ},110.96^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(34,43,-21)\\&\rho=\sqrt{(34)^2+(43)^2+(-21)^2}=58.7\\&\theta=arctan(\frac{43}{34})=51.67^{\circ}\\&\phi=arccos(\frac{-21}{58.7})=110.96^{\circ}\end{align*}$

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Example Question #21 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-19,-21,-13) . What is the position of this point in spherical coordinates?

Possible Answers:

$(28.32,47.86^{\circ},65.34^{\circ})$

$(28.32,-132.14^{\circ},114.66^{\circ})$

$(31.16,-132.14^{\circ},114.66^{\circ})$

$(28.32,47.86^{\circ},-65.34^{\circ})$

$(31.16,47.86^{\circ},114.66^{\circ})$

Correct answer:

$(31.16,-132.14^{\circ},114.66^{\circ})$

Explanation:

$\rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\theta$ , the formula for it is as follows:

$\theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both and , bearing in mind which quadrant the point lies; this will determine the value of $\theta$ :

Quadrants

To calculate $\phi$ , we can make use of our $\rho$ value. It is found as

$\phi=arccos(\frac{z}{\rho})$ , where $0\leq\phi\leq180^{\circ}$

Phirho

For our coordinates

$\begin{align*}&(-19,-21,-13)\\&\rho=\sqrt{(-19)^2+(-21)^2+(-13)^2}=31.16\\&\theta=arctan(\frac{-21}{-19})=-132.14^{\circ}\\&\phi=arccos(\frac{-13}{31.16})=114.66^{\circ}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #353 : 3 Dimensional Space

Example Question #354 : 3 Dimensional Space

Example Question #355 : 3 Dimensional Space

Example Question #356 : 3 Dimensional Space

Example Question #357 : 3 Dimensional Space

Example Question #358 : 3 Dimensional Space

Example Question #11 : Spherical Coordinates

Example Question #12 : Spherical Coordinates

Example Question #13 : Spherical Coordinates

Example Question #21 : Spherical Coordinates

All Calculus 3 Resources

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