Spherical Coordinates

Help Questions

AP Calculus BC › Spherical Coordinates

Questions 1 - 10

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

$(9.7,18.43^{\circ},78.1^{\circ})$

$(9.7,161.57^{\circ},78.1^{\circ})$

$(9.49,18.43^{\circ},78.1^{\circ})$

$(9.49,161.57^{\circ},78.1^{\circ})$

$(9.49,161.57^{\circ},-101.9^{\circ})$

Explanation

When given Cartesian coordinates of the form 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

$\begin{align*}&(9,3,2)\&\rho=\sqrt{(9)^2+(3)^2+(2)^2}=9.7\&\theta=arctan(\frac{3}{9})=18.43^{\circ}\&\phi=arccos(\frac{2}{9.7})=78.1^{\circ}\end{align*}$

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

$(41.77,-142.13^{\circ},33.09^{\circ})$

$(41.77,37.87^{\circ},33.09^{\circ})$

$(22.8,-142.13^{\circ},33.09^{\circ})$

$(22.8,37.87^{\circ},146.91^{\circ})$

$(22.8,37.87^{\circ},-146.91^{\circ})$

Explanation

When given Cartesian coordinates of the form 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

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

$\begin{align*}&(-18,-14,35)\&\rho=\sqrt{(-18)^2+(-14)^2+(35)^2}=41.77\&\theta=arctan(\frac{-14}{-18})=-142.13^{\circ}\&\phi=arccos(\frac{35}{41.77})=33.09^{\circ}\end{align*}$

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

$(51.62,-132.27^{\circ},59.76^{\circ})$

$(44.6,-132.27^{\circ},59.76^{\circ})$

$(44.6,47.73^{\circ},120.24^{\circ})$

$(44.6,47.73^{\circ},-120.24^{\circ})$

$(51.62,47.73^{\circ},59.76^{\circ})$

Explanation

When given Cartesian coordinates of the form 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

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

$\begin{align*}&(-30,-33,26)\&\rho=\sqrt{(-30)^2+(-33)^2+(26)^2}=51.62\&\theta=arctan(\frac{-33}{-30})=-132.27^{\circ}\&\phi=arccos(\frac{26}{51.62})=59.76^{\circ}\end{align*}$

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(35,-8,-45).\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

$(57.57,-12.88^{\circ},141.42^{\circ})$

$(57.57,167.12^{\circ},141.42^{\circ})$

$(36.53,-12.88^{\circ},141.42^{\circ})$

$(36.53,167.12^{\circ},38.58^{\circ})$

Explanation

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\& \text{calculate the }\rho\text{ term first:}\& \rho=\sqrt{x^2+y^2+z^2}\&\text{Next, calculate angles, being prudent when calculating them}.\&\text{The formula for }\theta\text{ is as follows:}\& \theta=arctan(\frac{y}{x})\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\& \text{which quadrant the point lies in.}\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\&\text{values lead to }|\theta|>90^{\circ}\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\&\text{For our coordinates: }(35,-8,-45)\&\rho=\sqrt{(35)^2+(-8)^2+(-45)^2}=57.57\&\theta=arctan(\frac{-8}{35})=-12.88^{\circ}\&\phi=arccos(\frac{-45}{57.57})=141.42^{\circ}\end{align*}$

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-67,14,138).\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

$(154.04,168.2^{\circ},26.38^{\circ})$

$(132.57,168.2^{\circ},26.38^{\circ})$

$(132.57,11.8^{\circ},153.62^{\circ})$

$(154.04,11.8^{\circ},26.38^{\circ})$

Explanation

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\& \text{calculate the }\rho\text{ term first:}\& \rho=\sqrt{x^2+y^2+z^2}\&\text{Next, calculate angles, being prudent when calculating them}.\&\text{The formula for }\theta\text{ is as follows:}\& \theta=arctan(\frac{y}{x})\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\& \text{which quadrant the point lies in.}\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\&\text{values lead to }|\theta|>90^{\circ}\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\&\text{For our coordinates: }(-67,14,138)\&\rho=\sqrt{(-67)^2+(14)^2+(138)^2}=154.04\&\theta=arctan(\frac{14}{-67})=168.2^{\circ}\&\phi=arccos(\frac{138}{154.04})=26.38^{\circ}\end{align*}$

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

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

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

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

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

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

Explanation

When given Cartesian coordinates of the form 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

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

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

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-32,52,73).\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

$(95.17,121.61^{\circ},39.91^{\circ})$

$(95.17,58.39^{\circ},39.91^{\circ})$

$(61.06,58.39^{\circ},-140.09^{\circ})$

$(61.06,58.39^{\circ},140.09^{\circ})$

$(61.06,121.61^{\circ},39.91^{\circ})$

Explanation

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\& \text{calculate the }\rho\text{ term first:}\& \rho=\sqrt{x^2+y^2+z^2}\&\text{Next, calculate angles, being prudent when calculating them}.\&\text{The formula for }\theta\text{ is as follows:}\& \theta=arctan(\frac{y}{x})\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\& \text{which quadrant the point lies in.}\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\&\text{values lead to }|\theta|>90^{\circ}\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\&\text{For our coordinates: }(-32,52,73)\&\rho=\sqrt{(-32)^2+(52)^2+(73)^2}=95.17\&\theta=arctan(\frac{52}{-32})=121.61^{\circ}\&\phi=arccos(\frac{73}{95.17})=39.91^{\circ}\end{align*}$

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(38,82,130).\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

$(158.33,65.14^{\circ},34.81^{\circ})$

$(158.33,114.86^{\circ},34.81^{\circ})$

$(90.38,65.14^{\circ},34.81^{\circ})$

$(90.38,114.86^{\circ},145.19^{\circ})$

$(90.38,114.86^{\circ},-145.19^{\circ})$

Explanation

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\& \text{calculate the }\rho\text{ term first:}\& \rho=\sqrt{x^2+y^2+z^2}\&\text{Next, calculate angles, being prudent when calculating them}.\&\text{The formula for }\theta\text{ is as follows:}\& \theta=arctan(\frac{y}{x})\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\& \text{which quadrant the point lies in.}\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\&\text{values lead to }|\theta|>90^{\circ}\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\&\text{For our coordinates: }(38,82,130)\&\rho=\sqrt{(38)^2+(82)^2+(130)^2}=158.33\&\theta=arctan(\frac{82}{38})=65.14^{\circ}\&\phi=arccos(\frac{130}{158.33})=34.81^{\circ}\end{align*}$

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(100,81,-100).\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

$(162.98,39.01^{\circ},127.85^{\circ})$

$(162.98,140.99^{\circ},127.85^{\circ})$

$(81,39.01^{\circ},127.85^{\circ})$

$(81,140.99^{\circ},52.15^{\circ})$

Explanation

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\& \text{calculate the }\rho\text{ term first:}\& \rho=\sqrt{x^2+y^2+z^2}\&\text{Next, calculate angles, being prudent when calculating them}.\&\text{The formula for }\theta\text{ is as follows:}\& \theta=arctan(\frac{y}{x})\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\& \text{which quadrant the point lies in.}\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\&\text{values lead to }|\theta|>90^{\circ}\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\&\text{For our coordinates: }(100,81,-100)\&\rho=\sqrt{(100)^2+(81)^2+(-100)^2}=162.98\&\theta=arctan(\frac{81}{100})=39.01^{\circ}\&\phi=arccos(\frac{-100}{162.98})=127.85^{\circ}\end{align*}$

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(6,-46,-105).\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

$(114.79,-82.57^{\circ},156.16^{\circ})$

$(114.79,97.43^{\circ},156.16^{\circ})$

$(46.39,-82.57^{\circ},156.16^{\circ})$

$(46.39,97.43^{\circ},23.84^{\circ})$

$(46.39,97.43^{\circ},-23.84^{\circ})$

Explanation

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\& \text{calculate the }\rho\text{ term first:}\& \rho=\sqrt{x^2+y^2+z^2}\&\text{Next, calculate angles, being prudent when calculating them}.\&\text{The formula for }\theta\text{ is as follows:}\& \theta=arctan(\frac{y}{x})\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\& \text{which quadrant the point lies in.}\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\&\text{values lead to }|\theta|>90^{\circ}\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\&\text{For our coordinates: }(6,-46,-105)\&\rho=\sqrt{(6)^2+(-46)^2+(-105)^2}=114.79\&\theta=arctan(\frac{-46}{6})=-82.57^{\circ}\&\phi=arccos(\frac{-105}{114.79})=156.16^{\circ}\end{align*}$

Page 1 of 18