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Use spherical coordinates I0 evaluate IIIvzav where E is the region that lies above the cone 0=1/3 and below the sphere p = |...

Question

Use spherical coordinates I0 evaluate IIIvzav where E is the region that lies above the cone 0=1/3 and below the sphere p = |

Use spherical coordinates I0 evaluate IIIvzav where E is the region that lies above the cone 0=1/3 and below the sphere p = |



Answers

Use spherical coordinates.
Evaluate $\iint_{E} y^{2} z^{2} d V,$ where $E$ lies above the cone $\phi=\pi / 3$
and below the sphere $\rho=1$

We're getting a great y squared Z squared over this um shape. So above the cone Row equals pi over 3? So it's a really fat cone. Um because power for three is 60°. So from here to there is 60° And below this fear row equals one. That's the green thing. Okay so row is starting here at zero and going up to the sphere. So row will be going from 0-1. Oops I meant to erase the mess I made there. Okay row equals zero to row equals one fee. Always starts here at the Z. Axis. So this is equal zero and then it's going to stop when it gets to fee equals pi over three because that's the answer to comb. So that's what's keeping it inside the cone this fee equals zero. Duffy Equals five or 3. And then tha tha well it's the whole cone. So data has to go all the way around to get the whole cone. So data 0 to 2 pi. Yeah. Okay so now I'm just gonna leave those hang in there. Let's see why squared. So why is rho sine fee sine theta. And that's squared. And then Z squared will be row co sign fee squared. And then devi is rho squared sign fee dear O. D. V. D. Theta. All right so here we have a row squared. A row squared in a row squared. So we have road to the 6th and we have a sign fee squared and a coastline fee squared. And another sign feast. So we have sign cube fee cosine squared fee. And then we have a sine squared data. Okay that's only one of those dear. O. D. V. D. Theme. Okay do you have the morning? Okay so we can go ahead and integrate row to the 6th dear. Oh and we'll get Road to the 7th over seven from 0 to one. Thank you go some squared sine squared. Do you feed the rope defeated data? Okay so that gives us 1/7. So let's put that out here. Yeah. All right now here I have a odd number of signs and an even number of co signs. So I'm going to take one of these signs and split it away from the other two. So I have sine squared fee, coastlines squared fee sign fee D. V. And then we still have the sine squared theta D. Theta. Yeah. Okay. The reason I did that because I want this to be D'you so I'm letting you be the co sign. And so d'you is minus the sign. So I need a minus sign in here which will put a minus sign out here and I need to replace the sine square. So now I have line is 17 integral integral 1- Cosine squared coastline squared minus sine squared minus sign. Sorry the fee. And this one was the was 0 to Pi over three. Okay if you equal zero I'm sorry if uh V. Equals zero then you equals the cosine of zero which is one If he equals pi over three Then you equals the cosine of pi over three. Uh huh. I have three square 321 Which goes 1/2. So this turns into 1- one half. One minus U. Squared U. Squared D. You. It's one half that's U. Squared minus U. To the fourth. D. You. Okay so you cubed over three minus you to the 5/5 from 1 to 1 half. So 1/8 over three -132/5. Mhm -1 3rd -1 5th. All right. I'm gonna work on that over here. I got one 24th minus one third minus 160th plus one fifth. So this needs an eight and an eight. So -7/24. And then this needs 32 minus plus 31/1 60. I can't Mhm. I'm just gonna do this minus seven times 1 60 plus 31 times 24 Over 24 times 1 60. And then I'm going to reduce instead of trying to sit here trying to think of what the common denominator is. 160 times seven so minus 11 20 plus 31 times 24 7 44. Why nestle. Okay, so now I have 376 Over 24 times 1 60. So I'm going to divide by 218, 8 Over 12 times 1 60 94. six times 1 60 47 over three times 1 60. I'm not liking that very much minus one. Seventh times 47/3 times 1 60. Hope that was negative. And now I have to integrate the sine squared of theta. D. Theta. So 47/7 times three times 1 60 I have to put in an identity one minus the co sign of two theta over to this data. So 47/7 times three times 1 60 times two Integral of one with respect to data is data-. Can we need a two in here to make the you and another one half. So one half signed your data We're going 0- two pi. So 47 times seven times three times 1 60 times two times two pi. Mhm. Yeah seven time oops seven times 3 times 1 60 16. 33 60. Dude. Yeah okay let's see if I can. Yeah yeah yeah okay so I paused the video I'm not sure what that looks like on your side. And I did I recalculated and I got minus 47 of 4 80 so I'm not sure where this seven came from. So we got 47 minus. Uh huh. Then somehow it turned oh yeah it's this 7:00 47/7 times for 80 times pi. That's what I get for the answer. It's kind of gross but I can't find anything else wrong

All right. We want to evaluate this triple integral above this cone in between these two spears. Um And it looks really, really ugly but it's not actually as bad as um You might think so uh One thing that makes it not bad is we look at this thing um And that's just row in spherical coordinates. It rose distance from the origin. So rho squared is X squared plus Y squared plus Z squared. And then another way way we can use that is that now these fears this is just really equals one and this is just really equals two. Rho squared equals one and four. So radical. Wanted to. Okay, that's good. Um And then this cone the cone is actually taken care of by fee. Because if we think about fee it's the angle with the positive Z. Axis and this cone it's super nice. This is actually just Z squared equals x squared plus y squared. Or if you think in polar coordinates you can think of it as Z squared equals R squared. Or jay Z equals R. So every time we increase Z we're drawing a circle of radius Z. And because it's Z equals R. With no numbers not Z equals two. Are anything. This angle that it makes is 45°. I think it's like it's like the line Y equals X. But it's the equals R. So then um that cone is just founded by fee equals 45° or pi over four. Okay, so we're above this cone which means fee is between zero and pi over four because we're basically inside it. And then we're between these spheres. Sofia so rose between one. We should make that look more like a ro ro is between one and two. And then um it's all the way around the cone. So that means data goes from 0-2 pi. So I can write that in. So 1-2 for row 0 to Pi over four for fee and 0- two pi for data. And then that was fee that was row. And then I have a rho squared sine fee for converting to polar coordinates. And then I have a dear oh D. V. D. Theater. And now I just need to perform that integration so that's um row cute. So if I integrate that I get road to the 4th over four Going from should make that more like a four less like a row going from 1-2. And then I leave everything else here signed the d uh sorry D. Fee. I'll write it like the other fee. So it looks like your textbook. Actually. All right this sci fi there's two different ways to write fee. You can write fee and fee. So I don't know if you have a professor with lectures but they might write it the other way. Your textbook will always write it this way they're both the greek letter fee. Um D theta. Okay so now we plug in pounds 0-2 pi 0 to Pi over 4. 2- 4 16. So that's four. Um Once you divide by 4 -5. Okay so that's 15 force sign. Fear D. Fi the theater we integrate sign fear we get negative co sign so negative 15/4 co sign fee going from zero to pi over four do you say to um Okay so zero two pi we're putting in a co sign of pi over four is one of a route to so negative. It's gonna be a little bit ugly for route to and then minus negative is a plus 15/4. Um and this is actually it doesn't matter because it's not a volume so it doesn't have to be positive but this is actually a positive number because this first number is smaller than the second number we're dividing by route to which is about 1.4. Okay the data and then this is just going to integrate to tha to going from 0 to 2 pi so we'll have a factor of two pi So I have 15 Pi over two. We'll just cancel out that extra to their and then um that'll be your answer. So don't be scared by spherical coordinates. Um It's usually not that bad. The immigration is usually turn out to be nice

And this question we're going to evaluate an integral in spherical coordinates given an inte grand as well as the region of integration. So with that being said, the first thing that we should do is we should remember our hysterical conversions. The first thing that we're going to do is we're going to notice that rho squared equals X squared plus y squared plus Z plus t squared. And we know that the differential volume is equal to rho squared times the sine of phi time's d rho d fada time zero decided to defy this is going to be important for later because of this. We can immediately convert the Inte Grant and the differential into an integral. So we know that it's going to be the integral over some region over some region E we'll have the square root of X squared plus one of y squared plus b squared. That's just going to be row. And of course the differential volume is rho squared times the sine of phi time's d rho d say to defy Yeah. Yeah. Now obviously the harder part is making the inte grand to finding out what our limits of integration our So we'll start here with the two spheres. If we're in between two spheres we know we're going to find out that we're going to be in between a radius of one and two. In between. We have row rho squared equals one and rose squared equals four. Which implies that row will be in between of one and two. Okay, so now that we know that we don't need to care about any negative values because it's just very awkward. Yeah. Now what are our fi limits? The five limits are a little bit harder to find out. So the five limits are going to involve the cone, The cone is this equation here Z equals the square root of x squared plus y squared. So we know that the cone. How do we deal with this? Well, we have to remember this spherical conversion for Z. We know that Z is equal to rho times the coastline of five. Yeah, so motivated by that. We can make use of our motivated by that. We can use some, we can do a little bit of algebra to this cone. For instance, if I square both sides, we'll get that. We have just X squared plus y squared. And then here I'm going to add, I'm going to add Z square to both sides. So we'll have to Z squared equals X squared plus y squared plus Z squared. But we know but we remember that extra plus Y squared plus Z squared equals rho squared. So we can substitute this in. And also substitute our definition for Z. When we do so, we're going to get that. We have two times rho squared times the cosine squared of phi is equal to rho squared And the roast where it's obviously vanished right? Which means that the co sign the coastline of Fi has to be equal to one hat whatever discredit of two by taking just a square root, which means that Fi has to be equal to pi over four. Now that will be our our upper bound for Fi. This is going to imply that Fi is going to be in between zero and pi over four, wow. And for the fate of bounds, we don't really need to we don't need to worry about that because we just know if nothing is stated, we know that we're going to assume that data is in between zero and two pi. So what that does for us, it gives us this integral and integral from zero to pi over four. An integral from 0 to 2 pi. And an integral from 1 to 2. And again we have rho cubed times the sine of phi zero d. Day. To defy Okay. And each of these can be integrated very easily because we can split this into a product of three integral. So this is going to be the integral from zero to pi over four. The sine of phi defy an integral from zero to pi ridge from 0 to 2 pi. We'll have one D theta and an integral from from 1 to 2. I've row cubed. Dear. Oh, okay, evaluating each of these is pretty straightforward because the integral of sine is negative the co sign and that will be integrated from zero to pi over four. The second integral is just two pi and the other integral is just 1/4 times wrote to the fourth power evaluated from 1 to 2. Now negative. The cosine of pi over four gives us all right. This will result in having negative square root of 2/2 minus one. He minus negative one, which is just plus one times two pi. All of that times two pi. All of those times 15/4. Yeah, which will simplify really nicely in the end to 15/4. Mhm times two minus the spirit of two times pi. And that is the answer to this question.

Were given a region and were asked to find the volume of this region. The region lies above the cone. Phi equals fine not and below the sphere Row equals big are well, we're really looking at on Lee, the upper hemisphere of this sphere. It's where a region is described with the inequalities. W is data lies between zero and two pi five lives between zero and pi over two. And this is because well, not actually Piper to zero and find not because we have to be above the cone and then roe ranges from zero up to the radius. Big are of the spear. We can compute the volume of w using triple into roles in spherical coordinates. So we have our volume is the triple integral over W of one which is the integral from 0 to 2 pi integral from zero to find Not integral from zero to big are of our function One times the differential roasts Weird signed fi dear Oh, defy the theta and using food beanies therapy, right? This is a product of integral. So we get the integral from 0 to 2 pi d theta times the integral from zero to find not sign Phi If I times integral from zero to big are of gross squared Dear oh, taking anti derivatives You get two pi times negative cosine phi from zero to find not times one third big are cute and so evaluating We get two pi times negative cosign of fine not plus co sign of one So we have two pi times one minus cosine of fine not times big r cubed all over three And this is our answer


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