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Calculus 3 : Spherical Coordinates

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Example Questions

Example Question #151 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(76,-36,20).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(86.44,154.65^{\circ},166.62^{\circ})$

$(86.44,154.65^{\circ},103.38^{\circ})$

$(86.44,-25.35^{\circ},13.38^{\circ})$

$(86.44,-25.35^{\circ},76.62^{\circ})$

Correct answer:

$(86.44,-25.35^{\circ},76.62^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(76,-36,20)\\&\rho=\sqrt{(76)^2+(-36)^2+(20)^2}=86.44\\&\theta=arctan(\frac{-36}{76})=-25.35^{\circ}\\&\phi=arccos(\frac{20}{86.44})=76.62^{\circ}\end{align*}$

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Example Question #152 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-128,-134,9).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(185.53,-133.69^{\circ},2.78^{\circ})$

$(185.53,46.31^{\circ},92.78^{\circ})$

$(185.53,-133.69^{\circ},87.22^{\circ})$

$(185.53,46.31^{\circ},177.22^{\circ})$

Correct answer:

$(185.53,-133.69^{\circ},87.22^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-128,-134,9)\\&\rho=\sqrt{(-128)^2+(-134)^2+(9)^2}=185.53\\&\theta=arctan(\frac{-134}{-128})=-133.69^{\circ}\\&\phi=arccos(\frac{9}{185.53})=87.22^{\circ}\end{align*}$

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Example Question #153 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(84,131,-111).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(191.15,-122.67^{\circ},54.5^{\circ})$

$(191.15,57.33^{\circ},125.5^{\circ})$

$(191.15,-122.67^{\circ},35.5^{\circ})$

$(191.15,57.33^{\circ},144.5^{\circ})$

Correct answer:

$(191.15,57.33^{\circ},125.5^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(84,131,-111)\\&\rho=\sqrt{(84)^2+(131)^2+(-111)^2}=191.15\\&\theta=arctan(\frac{131}{84})=57.33^{\circ}\\&\phi=arccos(\frac{-111}{191.15})=125.5^{\circ}\end{align*}$

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Example Question #154 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(21,-9,-147).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(148.76,-23.2^{\circ},171.17^{\circ})$

$(148.76,-23.2^{\circ},98.83^{\circ})$

$(148.76,156.8^{\circ},8.83^{\circ})$

$(148.76,156.8^{\circ},81.17^{\circ})$

Correct answer:

$(148.76,-23.2^{\circ},171.17^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(21,-9,-147)\\&\rho=\sqrt{(21)^2+(-9)^2+(-147)^2}=148.76\\&\theta=arctan(\frac{-9}{21})=-23.2^{\circ}\\&\phi=arccos(\frac{-147}{148.76})=171.17^{\circ}\end{align*}$

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Example Question #155 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-49,-102,89).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(143.97,64.34^{\circ},141.81^{\circ})$

$(143.97,-115.66^{\circ},38.19^{\circ})$

$(143.97,64.34^{\circ},128.19^{\circ})$

$(143.97,-115.66^{\circ},51.81^{\circ})$

Correct answer:

$(143.97,-115.66^{\circ},51.81^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-49,-102,89)\\&\rho=\sqrt{(-49)^2+(-102)^2+(89)^2}=143.97\\&\theta=arctan(\frac{-102}{-49})=-115.66^{\circ}\\&\phi=arccos(\frac{89}{143.97})=51.81^{\circ}\end{align*}$

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Example Question #156 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-57,9,-101).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(116.32,-8.97^{\circ},60.26^{\circ})$

$(116.32,-8.97^{\circ},29.74^{\circ})$

$(116.32,171.03^{\circ},119.74^{\circ})$

$(116.32,171.03^{\circ},150.26^{\circ})$

Correct answer:

$(116.32,171.03^{\circ},150.26^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-57,9,-101)\\&\rho=\sqrt{(-57)^2+(9)^2+(-101)^2}=116.32\\&\theta=arctan(\frac{9}{-57})=171.03^{\circ}\\&\phi=arccos(\frac{-101}{116.32})=150.26^{\circ}\end{align*}$

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Example Question #157 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(31,-71,46).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(90.1,113.59^{\circ},149.3^{\circ})$

$(90.1,-66.41^{\circ},30.7^{\circ})$

$(90.1,-66.41^{\circ},59.3^{\circ})$

$(90.1,113.59^{\circ},120.7^{\circ})$

Correct answer:

$(90.1,-66.41^{\circ},59.3^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(31,-71,46)\\&\rho=\sqrt{(31)^2+(-71)^2+(46)^2}=90.1\\&\theta=arctan(\frac{-71}{31})=-66.41^{\circ}\\&\phi=arccos(\frac{46}{90.1})=59.3^{\circ}\end{align*}$

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Example Question #501 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (19,-94^{\circ},33^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Possible Answers:

(18.42,4.66,-0.25)

(-0.24,-0.06,19)

(-0.72,-10.32,15.93)

(-1.11,-15.9,10.35)

Correct answer:

(-0.72,-10.32,15.93)

Explanation:

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=19sin(33^{\circ})cos(-94^{\circ})=-0.72\\&y=19sin(33^{\circ})sin(-94^{\circ})=-10.32\\&z=19cos(33^{\circ})15.93\end{align*}$

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Example Question #502 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (142,-4^{\circ},88^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Possible Answers:

(-3.29,3.8,141.91)

(141.57,-9.9,4.96)

(-92.76,107.4,5.03)

(4.94,-0.35,141.91)

Correct answer:

(141.57,-9.9,4.96)

Explanation:

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=142sin(88^{\circ})cos(-4^{\circ})=141.57\\&y=142sin(88^{\circ})sin(-4^{\circ})=-9.9\\&z=142cos(88^{\circ})4.96\end{align*}$

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Example Question #503 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (51,-47^{\circ},162^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Possible Answers:

(10.75,-11.53,-48.5)

(-10.45,-1.3,-49.9)

(-33.08,35.47,15.76)

(49.52,6.17,10.53)

Correct answer:

(10.75,-11.53,-48.5)

Explanation:

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=51sin(162^{\circ})cos(-47^{\circ})=10.75\\&y=51sin(162^{\circ})sin(-47^{\circ})=-11.53\\&z=51cos(162^{\circ})-48.5\end{align*}$

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Calculus 3 : Spherical Coordinates

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #151 : Spherical Coordinates

Example Question #152 : Spherical Coordinates

Example Question #153 : Spherical Coordinates

Example Question #154 : Spherical Coordinates

Example Question #155 : Spherical Coordinates

Example Question #156 : Spherical Coordinates

Example Question #157 : Spherical Coordinates

Example Question #501 : 3 Dimensional Space

Example Question #502 : 3 Dimensional Space

Example Question #503 : 3 Dimensional Space

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