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Calculus 3 : 3-Dimensional Space

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Example Questions

Example Question #401 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(142,-93,-109).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(169.74,-33.22^{\circ},122.71^{\circ})$

$(169.74,146.78^{\circ},-57.29^{\circ})$

$(201.73,146.78^{\circ},122.71^{\circ})$

$(201.73,-33.22^{\circ},122.71^{\circ})$

$(169.74,146.78^{\circ},57.29^{\circ})$

Correct answer:

$(201.73,-33.22^{\circ},122.71^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(142,-93,-109)\\&\rho=\sqrt{(142)^2+(-93)^2+(-109)^2}=201.73\\&\theta=arctan(\frac{-93}{142})=-33.22^{\circ}\\&\phi=arccos(\frac{-109}{201.73})=122.71^{\circ}\end{align*}$

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Example Question #402 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(59,-122,8).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(135.75,-64.19^{\circ},86.62^{\circ})$

$(135.75,115.81^{\circ},86.62^{\circ})$

$(135.52,115.81^{\circ},93.38^{\circ})$

$(135.52,-64.19^{\circ},86.62^{\circ})$

$(135.52,115.81^{\circ},-93.38^{\circ})$

Correct answer:

$(135.75,-64.19^{\circ},86.62^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(59,-122,8)\\&\rho=\sqrt{(59)^2+(-122)^2+(8)^2}=135.75\\&\theta=arctan(\frac{-122}{59})=-64.19^{\circ}\\&\phi=arccos(\frac{8}{135.75})=86.62^{\circ}\end{align*}$

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Example Question #51 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(9,109,-5).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(109.37,94.72^{\circ},87.38^{\circ})$

$(109.49,85.28^{\circ},92.62^{\circ})$

$(109.37,94.72^{\circ},-87.38^{\circ})$

$(109.37,85.28^{\circ},92.62^{\circ})$

$(109.49,94.72^{\circ},92.62^{\circ})$

Correct answer:

$(109.49,85.28^{\circ},92.62^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(9,109,-5)\\&\rho=\sqrt{(9)^2+(109)^2+(-5)^2}=109.49\\&\theta=arctan(\frac{109}{9})=85.28^{\circ}\\&\phi=arccos(\frac{-5}{109.49})=92.62^{\circ}\end{align*}$

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Example Question #403 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #404 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #405 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(26,-72,-137).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(156.94,-70.14^{\circ},150.81^{\circ})$

$(76.55,109.86^{\circ},29.19^{\circ})$

$(76.55,-70.14^{\circ},150.81^{\circ})$

$(156.94,109.86^{\circ},150.81^{\circ})$

Correct answer:

$(156.94,-70.14^{\circ},150.81^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(26,-72,-137)\\&\rho=\sqrt{(26)^2+(-72)^2+(-137)^2}=156.94\\&\theta=arctan(\frac{-72}{26})=-70.14^{\circ}\\&\phi=arccos(\frac{-137}{156.94})=150.81^{\circ}\end{align*}$

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Example Question #64 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(77,-77,-17).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(110.21,-45^{\circ},98.87^{\circ})$

$(108.89,135^{\circ},81.13^{\circ})$

$(110.21,135^{\circ},98.87^{\circ})$

$(108.89,-45^{\circ},98.87^{\circ})$

Correct answer:

$(110.21,-45^{\circ},98.87^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(77,-77,-17)\\&\rho=\sqrt{(77)^2+(-77)^2+(-17)^2}=110.21\\&\theta=arctan(\frac{-77}{77})=-45^{\circ}\\&\phi=arccos(\frac{-17}{110.21})=98.87^{\circ}\end{align*}$

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Example Question #406 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(57,-42,71).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(70.8,143.62^{\circ},135.08^{\circ})$

$(100.27,143.62^{\circ},44.92^{\circ})$

$(100.27,-36.38^{\circ},44.92^{\circ})$

$(70.8,-36.38^{\circ},44.92^{\circ})$

Correct answer:

$(100.27,-36.38^{\circ},44.92^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(57,-42,71)\\&\rho=\sqrt{(57)^2+(-42)^2+(71)^2}=100.27\\&\theta=arctan(\frac{-42}{57})=-36.38^{\circ}\\&\phi=arccos(\frac{71}{100.27})=44.92^{\circ}\end{align*}$

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Example Question #407 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-32,55,61).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(88.15,120.19^{\circ},46.21^{\circ})$

$(88.15,59.81^{\circ},46.21^{\circ})$

$(63.63,120.19^{\circ},46.21^{\circ})$

$(63.63,59.81^{\circ},133.79^{\circ})$

Correct answer:

$(88.15,120.19^{\circ},46.21^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-32,55,61)\\&\rho=\sqrt{(-32)^2+(55)^2+(61)^2}=88.15\\&\theta=arctan(\frac{55}{-32})=120.19^{\circ}\\&\phi=arccos(\frac{61}{88.15})=46.21^{\circ}\end{align*}$

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Example Question #408 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-17,-145,-51).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(154.64,83.31^{\circ},109.26^{\circ})$

$(145.99,83.31^{\circ},70.74^{\circ})$

$(145.99,-96.69^{\circ},109.26^{\circ})$

$(154.64,-96.69^{\circ},109.26^{\circ})$

Correct answer:

$(154.64,-96.69^{\circ},109.26^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-17,-145,-51)\\&\rho=\sqrt{(-17)^2+(-145)^2+(-51)^2}=154.64\\&\theta=arctan(\frac{-145}{-17})=-96.69^{\circ}\\&\phi=arccos(\frac{-51}{154.64})=109.26^{\circ}\end{align*}$

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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 #401 : 3 Dimensional Space

Example Question #402 : 3 Dimensional Space

Example Question #51 : Spherical Coordinates

Example Question #403 : 3 Dimensional Space

Example Question #404 : 3 Dimensional Space

Example Question #405 : 3 Dimensional Space

Example Question #64 : Spherical Coordinates

Example Question #406 : 3 Dimensional Space

Example Question #407 : 3 Dimensional Space

Example Question #408 : 3 Dimensional Space

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