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Calculus 3 : 3-Dimensional Space

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #31 : 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:

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

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

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

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

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

Correct answer:

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

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

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

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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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Example Question #381 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (48,48,2) . What is the position of this point in spherical coordinates?

Possible Answers:

$(67.88,45^{\circ},88.31^{\circ})$

$(67.88,135^{\circ},-91.69^{\circ})$

$(67.88,135^{\circ},91.69^{\circ})$

$(67.91,45^{\circ},88.31^{\circ})$

$(67.91,135^{\circ},88.31^{\circ})$

Correct answer:

$(67.91,45^{\circ},88.31^{\circ})$

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

$\begin{align*}&(48,48,2)\\&\rho=\sqrt{(48)^2+(48)^2+(2)^2}=67.91\\&\theta=arctan(\frac{48}{48})=45^{\circ}\\&\phi=arccos(\frac{2}{67.91})=88.31^{\circ}\end{align*}$

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Example Question #382 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (-45,28,29) . What is the position of this point in spherical coordinates?

Possible Answers:

$(60.42,31.89^{\circ},61.31^{\circ})$

$(60.42,148.11^{\circ},61.31^{\circ})$

$(53,148.11^{\circ},61.31^{\circ})$

$(53,31.89^{\circ},-118.69^{\circ})$

$(53,31.89^{\circ},118.69^{\circ})$

Correct answer:

$(60.42,148.11^{\circ},61.31^{\circ})$

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

$\begin{align*}&(-45,28,29)\\&\rho=\sqrt{(-45)^2+(28)^2+(29)^2}=60.42\\&\theta=arctan(\frac{28}{-45})=148.11^{\circ}\\&\phi=arccos(\frac{29}{60.42})=61.31^{\circ}\end{align*}$

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Example Question #383 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (0,-31,-24) . What is the position of this point in spherical coordinates?

Possible Answers:

$(31,90^{\circ},-52.25^{\circ})$

$(31,-90^{\circ},127.75^{\circ})$

$(39.2,-90^{\circ},127.75^{\circ})$

$(31,90^{\circ},52.25^{\circ})$

$(39.2,90^{\circ},127.75^{\circ})$

Correct answer:

$(39.2,-90^{\circ},127.75^{\circ})$

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

$\begin{align*}&(0,-31,-24)\\&\rho=\sqrt{(0)^2+(-31)^2+(-24)^2}=39.2\\&\theta=arctan(\frac{-31}{0})=-90^{\circ}\\&\phi=arccos(\frac{-24}{39.2})=127.75^{\circ}\end{align*}$

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Example Question #384 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (0,-20,-34) . What is the position of this point in spherical coordinates?

Possible Answers:

$(39.45,90^{\circ},149.53^{\circ})$

$(20,90^{\circ},30.47^{\circ})$

$(39.45,-90^{\circ},149.53^{\circ})$

$(20,90^{\circ},-30.47^{\circ})$

$(20,-90^{\circ},149.53^{\circ})$

Correct answer:

$(39.45,-90^{\circ},149.53^{\circ})$

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

$\begin{align*}&(0,-20,-34)\\&\rho=\sqrt{(0)^2+(-20)^2+(-34)^2}=39.45\\&\theta=arctan(\frac{-20}{0})=-90^{\circ}\\&\phi=arccos(\frac{-34}{39.45})=149.53^{\circ}\end{align*}$

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Example Question #385 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (-41,25,44) . What is the position of this point in spherical coordinates?

Possible Answers:

$(48.02,31.37^{\circ},132.5^{\circ})$

$(65.13,148.63^{\circ},47.5^{\circ})$

$(48.02,148.63^{\circ},47.5^{\circ})$

$(65.13,31.37^{\circ},47.5^{\circ})$

$(48.02,31.37^{\circ},-132.5^{\circ})$

Correct answer:

$(65.13,148.63^{\circ},47.5^{\circ})$

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

$\begin{align*}&(-41,25,44)\\&\rho=\sqrt{(-41)^2+(25)^2+(44)^2}=65.13\\&\theta=arctan(\frac{25}{-41})=148.63^{\circ}\\&\phi=arccos(\frac{44}{65.13})=47.5^{\circ}\end{align*}$

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Example Question #386 : 3 Dimensional Space

A point in space is located, in Cartesian coordinates, at (-46,5,-4) . What is the position of this point in spherical coordinates?

Possible Answers:

$(46.27,6.2^{\circ},85.06^{\circ})$

$(46.44,6.2^{\circ},94.94^{\circ})$

$(46.27,173.8^{\circ},94.94^{\circ})$

$(46.44,173.8^{\circ},94.94^{\circ})$

$(46.27,6.2^{\circ},-85.06^{\circ})$

Correct answer:

$(46.44,173.8^{\circ},94.94^{\circ})$

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

$\begin{align*}&(-46,5,-4)\\&\rho=\sqrt{(-46)^2+(5)^2+(-4)^2}=46.44\\&\theta=arctan(\frac{5}{-46})=173.8^{\circ}\\&\phi=arccos(\frac{-4}{46.44})=94.94^{\circ}\end{align*}$

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Calculus 3 : 3-Dimensional Space

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #31 : Spherical Coordinates

Example Question #382 : 3 Dimensional Space

Example Question #383 : 3 Dimensional Space

Example Question #381 : 3 Dimensional Space

Example Question #381 : 3 Dimensional Space

Example Question #382 : 3 Dimensional Space

Example Question #383 : 3 Dimensional Space

Example Question #384 : 3 Dimensional Space

Example Question #385 : 3 Dimensional Space

Example Question #386 : 3 Dimensional Space

All Calculus 3 Resources

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