MATH 251 Lecture 24

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Triple Integrals

Given a region ,

Example with Simplification

The integral can be split into a sum of two integrals:

The first integral is over an odd region with respect to x, so for every point on one side, there is an equal and opposite point on the other side, so the entire integral will evaluate to 0:

The function inside this integral does not depend on x or y, so we can multiply the remaining integral with respect to z by the lengths of the intervals along x and y:

This is easy to evaluate as normal:

Triple Integrals in Spherical Coordinates

Find the center of mass of a unit hemisphere: , . Density is a constant

Find the mass: ... that was easy.

Due to symmentry, .

Therefore, we only need to calculate

This integral would be easier to evaluate by converting to cylindrical (polar) coordinates since

Another Example

, is bounded by in the first octant.

Let's evaluate in the order

Find the bounds:

  • (plug in 0 for )
  • (plug in 0 for and )

And due to time constraints, we will not evaluate that integral.