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Evaluate the integral f Jr Ie dV where R is the region between the sphere of radius and the sphere of radius centered at the oriigin and above the x-Y plane_...

Question

Evaluate the integral f Jr Ie dV where R is the region between the sphere of radius and the sphere of radius centered at the oriigin and above the x-Y plane_

Evaluate the integral f Jr Ie dV where R is the region between the sphere of radius and the sphere of radius centered at the oriigin and above the x-Y plane_



Answers

Evaluate the following integrals. $\iint_{R} x^{2} y d A ; R$ is the region in quadrants 1 and 4 bounded by the semicircle of radius 4 centered at (0,0)

Hello there in this exercise we need to avoid the following interval Where the region R is the finest and the first quadrant. Then the circles of radios one and 2. So just to make you quick sketch of how is defined. This integral here we are going to have A circle of Radius one in the first quadrant 1- one. And then we're going to have a circle on the set. The first squadron of radius too. This is the region that were defined here are so you can observe that we can translate this into the integral as follows. So it's going to be the integral between zero and pi half to find the first the first quadrant And then the integral between one and 2 for the radius part. Then we're going to perform here a change of variable. We know that X is supposed to co sign of theater times are and why is equal to design of theater are So if we put this together in this equation here, what we obtain here is our square times the coastline. Square of theater minus sign a square. Theater times are the are the data that is this differential of area for polar coordinates. Okay so let's a group of things here. So First we have here the integral between zero and pi half with respect to theta. Then here I'm going to put what depends on our so here will it put co sign square of theta minus sign a square of theater times. The integral between one and to off are acute D R. Defeated. Okay then we can integrate this really easy so we obtain The integral between zero and pi and we're going to write this. We know the reason. African American identity And there's equals two Co sign of 32 mm. The integral of our cube is are 4/3 On the Interval 1 2. And this we need to integrate with respect to the theater after below it. In the integral here on the limits We have the integral between zero and two pi of co sign of the peter I'm here. This is equal to 15. Here's four 15 4th the theater we can take out this this constant value from the integral because it doesn't depend on theta. We're going to take it out. So this will be equal to 15/4. And the integral of cosine of theater. The integral of co sign to theatre is a cost to 1/2 sign of the of the theater Between zero and by how and this result into 15/8 But Zain times sine of pi minus Sign of zero and both of these are zero. That means that there's Integral is opposed to zero

So first we need Thio. Figure out what? This region? Ours. Okay, so our only looking at the first quadrant. But what I'm going to do something to draw both of these circles. Okay, One has a radius of one one has a radius of three. But I only need everything in this first quadrant. Hey, so I'm looking at this portion here. This is our are. All right. So writing this as an interim lying from zero to pie over too. Afrin. One, two, three. That's our our bounds. Good sign. R Squared R D r d Better remember that This guy here, we have the relationship that that equal to r squared. Okay, so the first thing that we do is we change our bounds by using any substitution. I got this one half a swell. So from zero to hire a team integral of sign is negative. Co sign the view When? One. Tonight D NATO one. Huh? Times God! Co sign nine minus co. Sign of one. Okay, uh, some of this into your pirate king and we can see that far data into room. Well, just give us a pie over two nine zero so In the end, our answers in the PI over four times co sign one minutes co sign of night and that's it.

So we want to find the integral of a function over the Domaine de So Art Domaine De in this case is a sphere with a radius of nine. So if we were to draw that to find our bounds, we have that X goes from nine to negative nine because we have a radius of nine in the Y direction. We have negative nine to nine. So in the X Y plane, we have this circle with a radius of nine and then in the Z direction we have negative 9 to 9 are bounds. So we have a sphere with a radius of nine. Spheres are very difficult to draw, so we have our sphere here with our radius of nine. So we're going to look for the bounds. So the equation for a circle is X squared, plus y squared equals the radius squared. So if we want to add 1/3 dimension to make that a sphere, we're going to use X squared plus y squared plus z squared. Is that third? Um, is that 1/3 dimension and that's going to be still equal to the radius squared. So if we're looking for the Z bounce. We're going to have Z squared plus y squared plus X squared equals 81. So we're going to subtract y squared and X squared from both sides. So we get Z squared equals 81 minus y squared minus X squared. And then we take the square root of both sides so we get Z equals positive or negative square root of 81 minus y squared minus X squared. So those are going to be our bounds in the Z direction? No, If we fix Z at zero, our equation now becomes why squared Plus X squared equals the radius squared of 81. Because we have the Z squared, but it's equal to zero. So it's going to just not count for this. So if you want to find the why, bounds we're gonna have y squared equals 81 minus x squared. Take the square root of both sides of the get why equals positive or negative Square root of 81 minus x squared. So those are our why bounds and then for X we have we're going to fix Z and why at zero so that we have X squared equals 81 and taking the square root We have X equals positive or negative nine. So those are our expounds. So if we were to write an integral we go, we have the least the most simple on the outside of the very of the integral. So we're gonna have our ex bounds on the outside from negative 9 to 9 because we found that our bounds were in negative nine and positive nine. Next, we're going to have our integral That has two variables, which is our Why integral. So we're gonna have our bounds air negative, square root of 81 minus X squared and positive square rate of 81 minus x square. So we're going to write those for our lower and upper bounds for the Y variable. And finally, we're going on the in on the innermost integral. We have our most complex with three with two variables. So we have negative square root of 81 minus y squared minus X squared, and our upper bound is positive. 81 minus y squared minus X squared, and we're going to have but as the integral of the function of X y z times well, with respect to Z, with respect to why and with respect to X, going


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