We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 3 : Spherical Coordinates

Study concepts, example questions & explanations for Calculus 3

Create An Account

All Calculus 3 Resources

6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2099 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(145,81,25).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(167.96,29.19^{\circ},81.44^{\circ})$

$(153.2,29.19^{\circ},81.44^{\circ})$

$(153.2,150.81^{\circ},98.56^{\circ})$

$(167.96,150.81^{\circ},81.44^{\circ})$

Correct answer:

$(167.96,29.19^{\circ},81.44^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(145,81,25)\\&\rho=\sqrt{(145)^2+(81)^2+(25)^2}=167.96\\&\theta=arctan(\frac{81}{145})=29.19^{\circ}\\&\phi=arccos(\frac{25}{167.96})=81.44^{\circ}\end{align*}$

Report an Error

Example Question #91 : Spherical Coordinates

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(129,24,-145).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(195.56,169.46^{\circ},137.86^{\circ})$

$(96.12,10.54^{\circ},137.86^{\circ})$

$(195.56,10.54^{\circ},137.86^{\circ})$

$(96.12,169.46^{\circ},42.14^{\circ})$

Correct answer:

$(195.56,10.54^{\circ},137.86^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(129,24,-145)\\&\rho=\sqrt{(129)^2+(24)^2+(-145)^2}=195.56\\&\theta=arctan(\frac{24}{129})=10.54^{\circ}\\&\phi=arccos(\frac{-145}{195.56})=137.86^{\circ}\end{align*}$

Report an Error

Example Question #432 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-114,109,-5).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(157.8,136.28^{\circ},91.82^{\circ})$

$(57.15,136.28^{\circ},91.82^{\circ})$

$(57.15,43.72^{\circ},88.18^{\circ})$

$(157.8,43.72^{\circ},91.82^{\circ})$

Correct answer:

$(157.8,136.28^{\circ},91.82^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-114,109,-5)\\&\rho=\sqrt{(-114)^2+(109)^2+(-5)^2}=157.8\\&\theta=arctan(\frac{109}{-114})=136.28^{\circ}\\&\phi=arccos(\frac{-5}{157.8})=91.82^{\circ}\end{align*}$

Report an Error

Example Question #433 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(104,-87,16).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(77.77,140.09^{\circ},96.73^{\circ})$

$(136.53,-39.91^{\circ},83.27^{\circ})$

$(77.77,-39.91^{\circ},83.27^{\circ})$

$(136.53,140.09^{\circ},83.27^{\circ})$

Correct answer:

$(136.53,-39.91^{\circ},83.27^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(104,-87,16)\\&\rho=\sqrt{(104)^2+(-87)^2+(16)^2}=136.53\\&\theta=arctan(\frac{-87}{104})=-39.91^{\circ}\\&\phi=arccos(\frac{16}{136.53})=83.27^{\circ}\end{align*}$

Report an Error

Example Question #434 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(39,-141,35).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(150.42,105.46^{\circ},76.55^{\circ})$

$(150.42,-74.54^{\circ},76.55^{\circ})$

$(139.27,-74.54^{\circ},76.55^{\circ})$

$(139.27,105.46^{\circ},103.45^{\circ})$

Correct answer:

$(150.42,-74.54^{\circ},76.55^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(39,-141,35)\\&\rho=\sqrt{(39)^2+(-141)^2+(35)^2}=150.42\\&\theta=arctan(\frac{-141}{39})=-74.54^{\circ}\\&\phi=arccos(\frac{35}{150.42})=76.55^{\circ}\end{align*}$

Report an Error

Example Question #2101 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-41,-136,-3).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(142.08,73.22^{\circ},91.21^{\circ})$

$(137.23,73.22^{\circ},88.79^{\circ})$

$(142.08,-106.78^{\circ},91.21^{\circ})$

$(137.23,-106.78^{\circ},91.21^{\circ})$

Correct answer:

$(142.08,-106.78^{\circ},91.21^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-41,-136,-3)\\&\rho=\sqrt{(-41)^2+(-136)^2+(-3)^2}=142.08\\&\theta=arctan(\frac{-136}{-41})=-106.78^{\circ}\\&\phi=arccos(\frac{-3}{142.08})=91.21^{\circ}\end{align*}$

Report an Error

Example Question #2102 : Calculus 3

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(-93,-113,-89).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(143.46,-129.45^{\circ},121.31^{\circ})$

$(171.29,-129.45^{\circ},121.31^{\circ})$

$(143.46,50.55^{\circ},58.69^{\circ})$

$(171.29,50.55^{\circ},121.31^{\circ})$

Correct answer:

$(171.29,-129.45^{\circ},121.31^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-93,-113,-89)\\&\rho=\sqrt{(-93)^2+(-113)^2+(-89)^2}=171.29\\&\theta=arctan(\frac{-113}{-93})=-129.45^{\circ}\\&\phi=arccos(\frac{-89}{171.29})=121.31^{\circ}\end{align*}$

Report an Error

Example Question #441 : 3 Dimensional Space

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

Report an Error

Example Question #442 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(100,26,15).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(104.41,165.43^{\circ},81.74^{\circ})$

$(100.69,165.43^{\circ},98.26^{\circ})$

$(104.41,14.57^{\circ},81.74^{\circ})$

$(100.69,14.57^{\circ},81.74^{\circ})$

Correct answer:

$(104.41,14.57^{\circ},81.74^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(100,26,15)\\&\rho=\sqrt{(100)^2+(26)^2+(15)^2}=104.41\\&\theta=arctan(\frac{26}{100})=14.57^{\circ}\\&\phi=arccos(\frac{15}{104.41})=81.74^{\circ}\end{align*}$

Report an Error

Example Question #443 : 3 Dimensional Space

$\begin{align*}&\text{A point in space is located, in Cartesian coordinates, at }(126,-64,77).\\&\text{What is the position of this point in spherical coordinates?}\end{align*}$

Possible Answers:

$(129.96,-26.93^{\circ},61.42^{\circ})$

$(160.94,153.07^{\circ},61.42^{\circ})$

$(160.94,-26.93^{\circ},61.42^{\circ})$

$(129.96,153.07^{\circ},118.58^{\circ})$

Correct answer:

$(160.94,-26.93^{\circ},61.42^{\circ})$

Explanation:

$\begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(126,-64,77)\\&\rho=\sqrt{(126)^2+(-64)^2+(77)^2}=160.94\\&\theta=arctan(\frac{-64}{126})=-26.93^{\circ}\\&\phi=arccos(\frac{77}{160.94})=61.42^{\circ}\end{align*}$

Report an Error

View Calculus 3 Tutors

Emma
Certified Tutor

University of South Carolina, Bachelor of Science, Physics. University of South Carolina-Columbia, Bachelor of Science, Psych...

View Calculus 3 Tutors

Emma
Certified Tutor

University of Indianapolis, Bachelor of Science, Mathematics.

View Calculus 3 Tutors

Asad
Certified Tutor

NED University of Engineering and Technology, Bachelor of Science, Electrical Engineering. University of Cincinnati-Main Camp...

All Calculus 3 Resources

6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Chemistry Tutors in Denver, LSAT Tutors in Philadelphia, Chemistry Tutors in Phoenix, GMAT Tutors in Washington DC, MCAT Tutors in Boston, Math Tutors in New York City, ISEE Tutors in Boston, English Tutors in Houston, Math Tutors in San Diego, LSAT Tutors in San Diego

Popular Courses & Classes

Spanish Courses & Classes in Chicago, SSAT Courses & Classes in Denver, GRE Courses & Classes in New York City, GMAT Courses & Classes in Seattle, GRE Courses & Classes in Chicago, LSAT Courses & Classes in Philadelphia, SAT Courses & Classes in Denver, ISEE Courses & Classes in New York City, ACT Courses & Classes in Seattle, MCAT Courses & Classes in Miami

Popular Test Prep

GRE Test Prep in Chicago, ISEE Test Prep in New York City, GRE Test Prep in New York City, ISEE Test Prep in Atlanta, SSAT Test Prep in Miami, MCAT Test Prep in Denver, MCAT Test Prep in Chicago, ACT Test Prep in Dallas Fort Worth, LSAT Test Prep in Denver, MCAT Test Prep in Houston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 3 : Spherical Coordinates

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #2099 : Calculus 3

Example Question #91 : Spherical Coordinates

Example Question #432 : 3 Dimensional Space

Example Question #433 : 3 Dimensional Space

Example Question #434 : 3 Dimensional Space

Example Question #2101 : Calculus 3

Example Question #2102 : Calculus 3

Example Question #441 : 3 Dimensional Space

Example Question #442 : 3 Dimensional Space

Example Question #443 : 3 Dimensional Space

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: