Evaluate the cylindrical coordinate integrals in Exercises 23-28 ∫0^2 π ∫0^1 ∫r^1 / √(2-r^2) 3 d

Evaluate the cylindrical coordinate integrals in Exercises...

Key Concepts

Order of Integration

The order of integration refers to the sequence in which the multiple integrals are evaluated. In cylindrical coordinates, the integration is often performed with respect to z, r, and then ?, though the order may vary based on the problem’s setup. Mastering the order of integration is important for simplifying the computation and ensuring that the integrals are evaluated correctly.

Integration Limits and Region Description

In the context of multiple integrals, the integration limits define the boundaries of the region over which the integration is performed. Analyzing these limits in cylindrical coordinates allows one to visualize the geometric region and appropriately set up the integral. This concept is essential for converting physical or geometric descriptions into precise mathematical limits for integration.

Cylindrical Coordinate System

This concept involves describing points in space using a set of coordinates (r, ?, z), where r is the distance from the z?axis, ? is the angle measured from a fixed direction in the xy-plane, and z is the height. It is especially useful for problems exhibiting rotational symmetry. Understanding cylindrical coordinates is essential for setting up and evaluating integrals over regions that are more naturally described in these terms.

Triple Integration

Triple integration is a method used to calculate volumes and other quantities over three-dimensional regions. It involves integrating a function of three variables sequentially over the three coordinate directions. This technique is crucial in multivariable calculus, particularly when dealing with regions in physical space.

Volume Element in Cylindrical Coordinates

When expressing triple integrals in cylindrical coordinates, the differential volume element is given by r dz dr d?. The factor r accounts for the Jacobian determinant that arises when converting from Cartesian to cylindrical coordinates, ensuring that area and volume are correctly scaled. Understanding this element is necessary for accurate integration in these coordinate systems.

Transcript

00:01 So you have the following integral in cylindrical coordinates, which are at least from 0 to 2 pi.

00:17 And from 0 up to 1, 1 between r and 1 over square root of 2 minus r squared.

00:37 We're square width of 2 minus r squared 3d z r d r d ther so this first bounds correspond to the bound on z the bound and this is obviously for there so for this first interval all the interval of z dc of 3 dc is equal to just 3 times z between the two bounds so that this will be equal to so 3 times this one c for the upper bound which is 1 over square root of 2 minus r squared minus 3 times the other bound are so that we have this r -tr between the next this is the next integral to do 0 and 1 and 0 to pi d theta so now all this integral can be separated into two intervals three r over square bit of two r squared and the interval of 3r squared well for this interval you can do a use substitution u equals to r squared u equals to r squared u equals to r squared so that this would be this integral 3 r d r t r squared square is of 2 my minus r squared while this is u and then the u would be equal to um to r d r so so that this would be equal to the integral of 3 over 2 because you move the d u half is equal to r d u over 2 d u 1 minus u so the integral of 1 minus u is equal to so 2 times 1 minus u and the way to check is well if we differentiate this we differentiate the right -hand side we should be at the left so if you differentiate that you have a minus 2 and then a half because this is here today one half and then that to the minus 1 half and then minus times a minus 1 which if you check is the same so this is the our interval and well this would be the integral of that so that putting it three halves so three halves of that is able to minus 3 halves and then squared that 0 .0b equal to minus 3 square root of 1 minus u and the r u is equal to r squared so that this interval the interval from 0 to 1 3 r d r r squared of 1 2 minus r squared will be equal to so we go to minus 3 times the square of 2 minus r squared but weight between 0 and 1 which is over 1 you have minus it was still 1 half minus minus 3 times 2 minus 1, 1 to 1 1 1 1 1 1.

06:58 And then at 0 minus 3 and then times the square root of 2.

07:13 So that this will be equal to 3 times the square root of 2 minus 1...

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