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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 #31 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (20,-9,2) . What is the position of this point in spherical coordinates?

Possible Answers:

$(21.93,-24.23^{\circ},84.79^{\circ})$

$(21.93,155.77^{\circ},95.21^{\circ})$

$(21.93,155.77^{\circ},-95.21^{\circ})$

$(22.02,155.77^{\circ},84.79^{\circ})$

$(22.02,-24.23^{\circ},84.79^{\circ})$

Correct answer:

$(22.02,-24.23^{\circ},84.79^{\circ})$

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

$\begin{align*}&(20,-9,2)\\&\rho=\sqrt{(20)^2+(-9)^2+(2)^2}=22.02\\&\theta=arctan(\frac{-9}{20})=-24.23^{\circ}\\&\phi=arccos(\frac{2}{22.02})=84.79^{\circ}\end{align*}$

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Example Question #32 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (19,24,26) . What is the position of this point in spherical coordinates?

Possible Answers:

$(40.16,51.63^{\circ},49.66^{\circ})$

$(30.61,128.37^{\circ},-130.34^{\circ})$

$(30.61,128.37^{\circ},130.34^{\circ})$

$(30.61,51.63^{\circ},49.66^{\circ})$

$(40.16,128.37^{\circ},49.66^{\circ})$

Correct answer:

$(40.16,51.63^{\circ},49.66^{\circ})$

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

$\begin{align*}&(19,24,26)\\&\rho=\sqrt{(19)^2+(24)^2+(26)^2}=40.16\\&\theta=arctan(\frac{24}{19})=51.63^{\circ}\\&\phi=arccos(\frac{26}{40.16})=49.66^{\circ}\end{align*}$

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Example Question #33 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (26,37,26) . What is the position of this point in spherical coordinates?

Possible Answers:

$(52.16,125.1^{\circ},60.1^{\circ})$

$(45.22,125.1^{\circ},-119.9^{\circ})$

$(45.22,54.9^{\circ},60.1^{\circ})$

$(52.16,54.9^{\circ},60.1^{\circ})$

$(45.22,125.1^{\circ},119.9^{\circ})$

Correct answer:

$(52.16,54.9^{\circ},60.1^{\circ})$

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

$\begin{align*}&(26,37,26)\\&\rho=\sqrt{(26)^2+(37)^2+(26)^2}=52.16\\&\theta=arctan(\frac{37}{26})=54.9^{\circ}\\&\phi=arccos(\frac{26}{52.16})=60.1^{\circ}\end{align*}$

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Example Question #35 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (23,36,50) . What is the position of this point in spherical coordinates?

Possible Answers:

$(42.72,57.43^{\circ},40.51^{\circ})$

$(42.72,122.57^{\circ},-139.49^{\circ})$

$(42.72,122.57^{\circ},139.49^{\circ})$

$(65.76,57.43^{\circ},40.51^{\circ})$

$(65.76,122.57^{\circ},40.51^{\circ})$

Correct answer:

$(65.76,57.43^{\circ},40.51^{\circ})$

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

$\begin{align*}&(23,36,50)\\&\rho=\sqrt{(23)^2+(36)^2+(50)^2}=65.76\\&\theta=arctan(\frac{36}{23})=57.43^{\circ}\\&\phi=arccos(\frac{50}{65.76})=40.51^{\circ}\end{align*}$

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Example Question #36 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (38,8,9) . What is the position of this point in spherical coordinates?

Possible Answers:

$(38.83,11.89^{\circ},76.95^{\circ})$

$(38.83,168.11^{\circ},-103.05^{\circ})$

$(38.83,168.11^{\circ},103.05^{\circ})$

$(39.86,168.11^{\circ},76.95^{\circ})$

$(39.86,11.89^{\circ},76.95^{\circ})$

Correct answer:

$(39.86,11.89^{\circ},76.95^{\circ})$

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

$\begin{align*}&(38,8,9)\\&\rho=\sqrt{(38)^2+(8)^2+(9)^2}=39.86\\&\theta=arctan(\frac{8}{38})=11.89^{\circ}\\&\phi=arccos(\frac{9}{39.86})=76.95^{\circ}\end{align*}$

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Example Question #34 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (-21,21,29) . What is the position of this point in spherical coordinates?

Possible Answers:

$(41.51,135^{\circ},45.68^{\circ})$

$(29.7,-45^{\circ},134.32^{\circ})$

$(29.7,-45^{\circ},-134.32^{\circ})$

$(41.51,-45^{\circ},45.68^{\circ})$

$(29.7,135^{\circ},45.68^{\circ})$

Correct answer:

$(41.51,135^{\circ},45.68^{\circ})$

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

$\begin{align*}&(-21,21,29)\\&\rho=\sqrt{(-21)^2+(21)^2+(29)^2}=41.51\\&\theta=arctan(\frac{21}{-21})=135^{\circ}\\&\phi=arccos(\frac{29}{41.51})=45.68^{\circ}\end{align*}$

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Example Question #35 : Spherical Coordinates

A point in space is located, in Cartesian coordinates, at (31,37,25) . What is the position of this point in spherical coordinates?

Possible Answers:

$(48.27,129.96^{\circ},-117.38^{\circ})$

$(54.36,129.96^{\circ},62.62^{\circ})$

$(48.27,129.96^{\circ},117.38^{\circ})$

$(54.36,50.04^{\circ},62.62^{\circ})$

$(48.27,50.04^{\circ},62.62^{\circ})$

Correct answer:

$(54.36,50.04^{\circ},62.62^{\circ})$

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

$\begin{align*}&(31,37,25)\\&\rho=\sqrt{(31)^2+(37)^2+(25)^2}=54.36\\&\theta=arctan(\frac{37}{31})=50.04^{\circ}\\&\phi=arccos(\frac{25}{54.36})=62.62^{\circ}\end{align*}$

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Example Question #382 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (-32,5,-29) . What is the position of this point in spherical coordinates?

Possible Answers:

$(43.47,8.88^{\circ},131.84^{\circ})$

$(43.47,171.12^{\circ},131.84^{\circ})$

$(32.39,171.12^{\circ},131.84^{\circ})$

$(32.39,8.88^{\circ},48.16^{\circ})$

$(32.39,8.88^{\circ},-48.16^{\circ})$

Correct answer:

$(43.47,171.12^{\circ},131.84^{\circ})$

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

$\begin{align*}&(-32,5,-29)\\&\rho=\sqrt{(-32)^2+(5)^2+(-29)^2}=43.47\\&\theta=arctan(\frac{5}{-32})=171.12^{\circ}\\&\phi=arccos(\frac{-29}{43.47})=131.84^{\circ}\end{align*}$

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Example Question #383 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (-18,2,13) . What is the position of this point in spherical coordinates?

Possible Answers:

$(18.11,6.34^{\circ},-125.67^{\circ})$

$(22.29,173.66^{\circ},54.33^{\circ})$

$(18.11,6.34^{\circ},125.67^{\circ})$

$(22.29,6.34^{\circ},54.33^{\circ})$

$(18.11,173.66^{\circ},54.33^{\circ})$

Correct answer:

$(22.29,173.66^{\circ},54.33^{\circ})$

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

$\begin{align*}&(-18,2,13)\\&\rho=\sqrt{(-18)^2+(2)^2+(13)^2}=22.29\\&\theta=arctan(\frac{2}{-18})=173.66^{\circ}\\&\phi=arccos(\frac{13}{22.29})=54.33^{\circ}\end{align*}$

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Example Question #381 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (-34,-12,-8) . What is the position of this point in spherical coordinates?

Possible Answers:

$(36.93,19.44^{\circ},102.51^{\circ})$

$(36.06,-160.56^{\circ},102.51^{\circ})$

$(36.06,19.44^{\circ},77.49^{\circ})$

$(36.06,19.44^{\circ},-77.49^{\circ})$

$(36.93,-160.56^{\circ},102.51^{\circ})$

Correct answer:

$(36.93,-160.56^{\circ},102.51^{\circ})$

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

$\begin{align*}&(-34,-12,-8)\\&\rho=\sqrt{(-34)^2+(-12)^2+(-8)^2}=36.93\\&\theta=arctan(\frac{-12}{-34})=-160.56^{\circ}\\&\phi=arccos(\frac{-8}{36.93})=102.51^{\circ}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #31 : Spherical Coordinates

Example Question #32 : Spherical Coordinates

Example Question #33 : Spherical Coordinates

Example Question #35 : Spherical Coordinates

Example Question #36 : Spherical Coordinates

Example Question #34 : Spherical Coordinates

Example Question #35 : Spherical Coordinates

Example Question #382 : 3 Dimensional Space

Example Question #383 : 3 Dimensional Space

Example Question #381 : 3 Dimensional Space

All Calculus 3 Resources

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