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Find the average value of the function $f(r, \theta, z)=r$ over the region bounded by the cylinder $r=1$ between the planes $z=-1$ and $z=1$

$\frac{2}{3}$

Calculus 3

Chapter 15

Multiple Integrals

Section 7

Triple Integrals in Cylindrical and Spherical Coordinates

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Lectures

04:18

In mathematics, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. Integration is one of the two main operations of calculus, with its inverse operation, differentiation, being the other. The area above the x-axis adds to the total.

26:18

In mathematics, a double integral is an integral where the integrand is a function of two variables, and the integral is taken over some region in the Euclidean plane.

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Find the average value of …

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Write an integral for the …

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Right so we want to find the average of this function. F of r theta c equals to r the average over the ceiling there ceiling there as radius 1 between the planes equals minus 1 and c gas 1 point. So this is a cylinder to describe it in cylindrical coordinates. We have here see a circle of radios 1 in the xpiplanewhen extended from 1 long to minus 1. So you have this vision. D want to find the average. So this is a cylinder it. It has a top area area is going to be. The area of a circle. Where is 1 is going to be his length there, for the area is going to be pi times 1 square, which is the radius that is bi, and this length here is 2. So that the volume of the is going to be equal to 2 pi the length times the area of the base, and so the average average is going to be equal to 1 over the volume of this region over being drawn over v of this function, the D x, d y d, but this function is for we have the function. R d x, d y d c would turn into r d c v: r theta in cylindrical, cylindrical, coordinates and so all here for we have but zis between minus 1 and 1 to go nap. That r goes from 0 up to 1 pagosa. The way around 0 to pi a whole 100 to pi and so we'll be doing this integral and then so. First, the integral of this there is discontent. Respect to this integral is going to be equal to z, below that 1 and minus 1 is equal to 1 minus minus 1, and this is equal to us. 1 plus 1 is to not 2 there at all be integral. This is just a just the space. So that this will be equal to 2 is equal to 2 times the integral from 0 up to 2 pi in trail from 0 up to 1 of our square, a r e theta, so central or squared yards arctis that evaluated between 0 and 1 will be Equal to that, which is 1 third minus 0, so that we have a factor of a third, a in 2 thirds gone from 0 to pi. It is than to her. The drop of theta is going to be theta between 2 pi and 0, which is equal to 2 pi minus 0 so that his going to be equal to 2 times 2 pi, which is that 2 pi divided by 3 and the equal to 1 over the Volume times this integral at this integral is that so that for the hours lone average is 1 equal to 1 over 2 pi to ply the volume times this integral over. Here it is equal to that times, 2 times 2 pithan, these 2 pi counsel. There we have that the average is going to be equal to 2 thirds, so that is the average of this function on the ceiling there.

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