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Calculus 3 : 3-Dimensional Space

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 #117 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-137,-121,97).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(206.93,-138.55^{\circ},62.05^{\circ})$

$(206.93,41.45^{\circ},117.95^{\circ})$

$(150.81,-138.55^{\circ},62.05^{\circ})$

$(150.81,41.45^{\circ},117.95^{\circ})$

Correct answer:

$(206.93,-138.55^{\circ},62.05^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-137,-121,97)\\&\rho=\sqrt{(-137)^2+(-121)^2+(97)^2}=206.93\\&\theta=arctan(\frac{-121}{-137})=-138.55^{\circ}\\&\phi=arccos(\frac{97}{206.93})=62.05^{\circ}\end{align*}$

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Example Question #118 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(59,-55,136).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(136.7,137.01^{\circ},149.33^{\circ})$

$(136.7,-42.99^{\circ},30.67^{\circ})$

$(158.12,-42.99^{\circ},30.67^{\circ})$

$(158.12,137.01^{\circ},149.33^{\circ})$

Correct answer:

$(158.12,-42.99^{\circ},30.67^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(59,-55,136)\\&\rho=\sqrt{(59)^2+(-55)^2+(136)^2}=158.12\\&\theta=arctan(\frac{-55}{59})=-42.99^{\circ}\\&\phi=arccos(\frac{136}{158.12})=30.67^{\circ}\end{align*}$

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Example Question #119 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-140,-18,-36).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(140.89,7.33^{\circ},75.69^{\circ})$

$(140.89,-172.67^{\circ},104.31^{\circ})$

$(145.67,-172.67^{\circ},104.31^{\circ})$

$(145.67,7.33^{\circ},75.69^{\circ})$

Correct answer:

$(145.67,-172.67^{\circ},104.31^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-140,-18,-36)\\&\rho=\sqrt{(-140)^2+(-18)^2+(-36)^2}=145.67\\&\theta=arctan(\frac{-18}{-140})=-172.67^{\circ}\\&\phi=arccos(\frac{-36}{145.67})=104.31^{\circ}\end{align*}$

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Example Question #461 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(80,89,-94).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(72.83,131.95^{\circ},51.85^{\circ})$

$(152.17,131.95^{\circ},51.85^{\circ})$

$(72.83,48.05^{\circ},128.15^{\circ})$

$(152.17,48.05^{\circ},128.15^{\circ})$

Correct answer:

$(152.17,48.05^{\circ},128.15^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(80,89,-94)\\&\rho=\sqrt{(80)^2+(89)^2+(-94)^2}=152.17\\&\theta=arctan(\frac{89}{80})=48.05^{\circ}\\&\phi=arccos(\frac{-94}{152.17})=128.15^{\circ}\end{align*}$

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Example Question #2131 : Calculus 3

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (121,148^{\circ},78^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Possible Answers:

(97.67,35.12,62.19)

(-100.37,62.72,25.16)

(-21.33,13.33,118.36)

(-58.52,-21.04,-103.79)

Correct answer:

(-100.37,62.72,25.16)

Explanation:

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=121sin(78^{\circ})cos(148^{\circ})=-100.37\\&y=118.356sin(78^{\circ})sin(148^{\circ})=62.72\\&z=121cos(78^{\circ})25.16\end{align*}$ a

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Example Question #2132 : Calculus 3

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (28,-128^{\circ},47^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Possible Answers:

(-2.4,-2.49,-27.79)

(-12.61,-16.14,19.1)

(19.25,20.03,3.46)

(-11.76,-15.05,20.48)

Correct answer:

(-12.61,-16.14,19.1)

Explanation:

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=28sin(47^{\circ})cos(-128^{\circ})=-12.61\\&y=20.4779sin(47^{\circ})sin(-128^{\circ})=-16.14\\&z=28cos(47^{\circ})19.1\end{align*}$

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Example Question #2133 : Calculus 3

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (21,29^{\circ},157^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Possible Answers:

(-16.91,-9.37,8.21)

(7.18,3.98,-19.33)

(-15.66,-13.89,-1.67)

(1.25,1.11,20.93)

Correct answer:

(7.18,3.98,-19.33)

Explanation:

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=21sin(157^{\circ})cos(29^{\circ})=7.18\\&y=8.20535sin(157^{\circ})sin(29^{\circ})=3.98\\&z=21cos(157^{\circ})-19.33\end{align*}$

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Example Question #2131 : Calculus 3

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (83,127^{\circ},26^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Possible Answers:

(-21.9,29.06,74.6)

(14.71,61.56,53.69)

(-44.9,59.58,36.38)

(12.48,52.22,63.29)

Correct answer:

(-21.9,29.06,74.6)

Explanation:

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=83sin(26^{\circ})cos(127^{\circ})=-21.9\\&y=36.3848sin(26^{\circ})sin(127^{\circ})=29.06\\&z=83cos(26^{\circ})74.6\end{align*}$

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Example Question #125 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (94,5^{\circ},63^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Possible Answers:

(83.44,7.3,42.68)

(42.51,3.72,83.75)

(26.29,-88.87,15.73)

(4.46,-15.09,92.67)

Correct answer:

(83.44,7.3,42.68)

Explanation:

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=94sin(63^{\circ})cos(5^{\circ})=83.44\\&y=83.7546sin(63^{\circ})sin(5^{\circ})=7.3\\&z=94cos(63^{\circ})42.68\end{align*}$

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Example Question #126 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in spherical coordinates, at } (61,-94^{\circ},13^{\circ})\text{ of form: } (\rho,\theta,\phi)\\&\text{Where }\phi\text{ is the angle between the vector, }\rho\text{, to the point and the }z\text{ axis}.\\&\text{What is the position of this point in Cartesian coordinates?}\end{align*}$

Possible Answers:

(-0.96,-13.69,59.44)

(-4.15,-59.29,13.72)

(24.85,6.29,55.35)

(53.66,13.58,25.63)

Correct answer:

(-0.96,-13.69,59.44)

Explanation:

$\begin{align*}&\text{When converting spherical coordinates of the form} (\rho,\theta,\phi)\\&\text{to Cartesian coordinates of the form }(x,y,z)\text{, it is necessary to relate } x\text{, }y\text{, and }z\\& \text{to the vector, }\rho\text{, and the angles }\theta.\text{ and }\phi.\text{ The relationships are as follows:}\\&x=\rho sin(\phi)cos(\theta)\\&y=\rho sin(\phi)sin(\theta)\\&z=\rho cos(\phi)\\&\text{You may notice that }\rho sin(\phi)\text{ is the projection of }\rho\text{ on the }xy\text{-plane!}\\&\text{However, care should be taken when finding these terms}\text{; if using a calculator, it is}\\& \text{imperative that the correct units (degrees or radians) are specified}\\& \text{for the input!}\\&\text{For our coordinates}\\&x=61sin(13^{\circ})cos(-94^{\circ})=-0.96\\&y=13.722sin(13^{\circ})sin(-94^{\circ})=-13.69\\&z=61cos(13^{\circ})59.44\end{align*}$

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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 #117 : Spherical Coordinates

Example Question #118 : Spherical Coordinates

Example Question #119 : Spherical Coordinates

Example Question #461 : 3 Dimensional Space

Example Question #2131 : Calculus 3

Example Question #2132 : Calculus 3

Example Question #2133 : Calculus 3

Example Question #2131 : Calculus 3

Example Question #125 : Spherical Coordinates

Example Question #126 : Spherical Coordinates

All Calculus 3 Resources

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