Iam not suggesting to make both latitude and longitude 360.My question is,stevendaryl said:Think about the Earth using latitude and longitude. Latitude runs from 90 degrees north latitude (the North Pole) to 90 degrees south. So that's 180 degrees. Longitude runs from 0 to 180 East longitude and then from 180 West longitude back to 0. So that's 360 degrees.
If you let latitude run a full 360 degrees, then points on Earth would have multiple coordinates: The point that is 270 degrees south of the North Pole and longitude 0 is the same as the point that is only 90 degrees south of the North Pole and longitude 180.
There is no problem doing that. However, the volume element is then different: you can calculate it using classical geometry or using the Jacobian of your new coordinate system. If you integrate the old volume element using your new coordinate system, what you are integrating will no longer represent volume.Mohankpvk said:Iam not suggesting to make both latitude and longitude 360.My question is,
why shouldn't we let latitude run 360 and make longitude 180?
Just inverting the system.
Thank you.Nice answer.slider142 said:There is no problem doing that. However, the volume element is then different: you can calculate it using classical geometry or using the Jacobian of your new coordinate system. If you integrate the old volume element using your new coordinate system, what you are integrating will no longer represent volume.
In particular, recall that the ##r\sin \phi ## factor of the volume element ##r^2 \sin\phi \, dr\, d\theta\, d\phi ## comes about geometrically from the length of the projection of the radial position vector of each point onto the xy-plane in order to get the sweeping radii for the lengths of arc ##d\theta ## and ##d\phi ##. If you are now allowing ##\phi ## to have values above 180°, then ##r\sin \phi ## will be a negative number when ##180^{\circ} < \phi < 360^{\circ}##. We do not want those projection lengths to be negative, so that is not the standard volume element anymore.
To get the correct standard volume, you can use the factor ##r|\sin \phi |## instead, which would result in a volume element of ##r^2 |\sin\phi | \, dr\, d\theta\, d\phi ##. However, I think you can see that this is a bit unwieldy.
Definitely. I do not know many good free books that go in depth the way that introductory books should, so unfortunately you will have to either pay for these texts or find them at a local library.Mohankpvk said:Thank you.Nice answer.
Is there any good(and simple,introductory) books for calculus and vector calculus?
Thank youslider142 said:Definitely. I do not know many good free books that go in depth the way that introductory books should, so unfortunately you will have to either pay for these texts or find them at a local library.
The best introductory textbook path I know to calculus and vector calculus is to first go through "Calculus" by Spivak, then study the texts "Vector Calculus, Linear Algebra, and Differential Forms" by Hubbard, "Calculus on Manifolds" by Spivak, "Linear Algebra Done Right" by Axler, and "Linear Algebra Done Wrong" by Treil concurrently.
The first is a nice walkthrough introduction of vector calculus that showcases many applications as well as showing details of vital theorems. The second gives an extremely bird's eye view of vector calculus that takes you from the basic component-based vector approach to the modern exterior calculus coordinate-independent approach with proper proofs and rigor. The latter two books help fill out the rigor in your understanding of linear algebra, as much of vector calculus is concerned with reducing nonlinear problems to questions in linear algebra. Knowing how linear algebra works properly is thus an important skill. Hope this helps you out! :-)
Spherical coordinates are a type of coordinate system used in three-dimensional space to describe the position of a point. They consist of three coordinates: radius (r), inclination angle (θ), and azimuth angle (φ).
In spherical coordinates, the triple integral is calculated using the volume element r2sinθdrdθdφ, whereas in Cartesian coordinates it is calculated using dx dy dz. This is because the volume element in spherical coordinates takes into account the curvature of the surface, while in Cartesian coordinates it is simply a rectangular prism.
Spherical coordinates are particularly useful when integrating over a region that is symmetric about the origin or when the boundaries of the region are defined by spheres or cones. They also simplify the calculation of volume elements for curved surfaces.
The conversion from Cartesian coordinates (x,y,z) to spherical coordinates (r,θ,φ) is done using the following equations:
r = √(x2 + y2 + z2)
θ = arccos(z/r)
φ = arctan(y/x)
No, spherical coordinates are primarily used in triple integrals to calculate volume, but they can also be used in double integrals to calculate surface area. In other types of integration, such as line integrals or surface integrals, other coordinate systems may be more appropriate.