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2. Use a double integral to compute the volume of a sphere centered at (0,0,0) and radius 7 in the JVa"_ x_y dy first octant (Hint: to compute use y=Va" ...

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2. Use a double integral to compute the volume of a sphere centered at (0,0,0) and radius 7 in the JVa"_ x_y dy first octant (Hint: to compute use y=Va" 'sin(0) and cos? 1+ cos (2 0) (0)=_ Also you may use polar coordinates to compute the double integral.)

2. Use a double integral to compute the volume of a sphere centered at (0,0,0) and radius 7 in the JVa"_ x_y dy first octant (Hint: to compute use y=Va" 'sin(0) and cos? 1+ cos (2 0) (0)=_ Also you may use polar coordinates to compute the double integral.)



Answers

(a) find the spherical coordinate limits for the integral that calculates the volume of the given solid and then (b) evaluate the integral. The solid bounded below by the sphere $\rho=2 \cos \phi$ and above by the cone $z=\sqrt{x^{2}+y^{2}}$

I think so. In this problem, we're going to be solving the equation, which we have a d a. So that most likely normally sense for area. And in this case, we have this area bound in the first coordinate, and it's above boundary is this circle, and it's boundary from below. Is this line so we can draw this out just to see what this was exactly look like? So this circle this a bead two. We have one. It's gonna be like this on our line is, like, said a bit more like that. Um, so it's this region. Okay, now from here, we're gonna do is we need Teoh first. I'm gonna write out, actually, um, three helpful questions so X is equal to our co sign data. Why, it's equal to our science data and R squared is equal to exported plus y squared. So first thing you can do is start trying to solve for r R. Um, and we're going to take our equation up there. The x minus one squared plus why squared is equal to one, and we're going to start plugging in the values for X and for why? So we have our coasts. It Ah, minus one. All this squared plus are stop sign data. All that squared is equal to one. This isn't the r squared coasts squared minus to our coasts. They, uh, plus one plus r squared sine squared data is equal to one so we can subtract are once and we get rid of that. And so when I say this is equal to zero, but just bringing the don't go this way. Um, so we're gonna have our squared coast signing squared, beta plus r squared, sine squared, beta minus two, our coast. I'm gonna move that last part. The other side's We're gonna add that both sides So what? Ever Plus it here. So that's going to give us to our close. They equal to r squared, Closest squared plus R squared sine squared bird. Okay. And now from here, what we can dio is divide out and are with my daughter. Arm would have to Coast data is equal to our coasts. Square data plus are signed squared data. And now we can take out our are gonna have coasts. But the plus sign squared data, and this is just equal to one So we're left with coast to coast. Data is equal toe are so that's one of our values for our right there. Andi Now, from here we're going to do is, uh I'm working out actually kind of solve for our data. So we have X is equal to y and R X is equal to our coasts data and our why is equal toe our sign data so we can divide. Um Why there are and cross out are ours and we're left with Kosaka and data is equal to sign Vada. And we know that that is true when Ada is equal to hide over for in the first coordinate. And so we have, um right now, two values. Um and what we're also going to do is because we're in the first coordinate. We're going to say that our fada is equal. Teoh hi. Over to the information that we have have been equal to hi over to It's in the first quarter and it's in the proportion First coordinate and we're going to have our r is equal to zero. And now we need to take our original equation up here and we're going to have to put that in polar coordinates that we have to. Why? And why is R sign fate? Us. That's all we need. And now we have our creation that we can put into a double integral with all of our polar values. So we're going to have from hi over or too high over to and for zero to two coasts sita of to our sign data. And then we have our d a, which is just going to be R D r de data. And now we're just going to separate our parts that have oh, are on with them in brackets and can start solving are into girl. So we're gonna have to r squared t r. And we'll have our sign data over here. Okay, And now the integral of Are squared is equal to are cute, so it's going to be our cubed over three. So we're gonna have to our cubed over three from 0 to 2 coasts Sita on this is times sign Data de data and we have the integral from however, four to I over to, and now we can start putting in our values. So from pi over two of and we're gonna take our to, um, Earth's out to make this easier. So of our cube. So in this case, we have to co sign data cubed minus zero cubed. This is a B times sign 70 data. This city is equal to zero. So that is going to get crossed out. So we're going to now have our Andi. We can cube too. So two cubed with eight that we're gonna have co sign cubed sign data. BP data on. We could take out our 8/3 on driver for two. High over to. And now from here, we're going to do IHS, um, we're gonna have co sign as you we're going to do a u substitution with Our co sons were going to say when you do it over here, So you is equal to co sign data. So be you is equal to negative Stein data, be it Seita. So do you. Over a negative sign. Data is equal to D data and I are gonna plug in. So we're gonna have Are you cute? Sign Saito. Times are e u. Negative sign. Data you can cross out are in betas and we can take our negative to the on. We can also start now solving for our introduced. The integral of you Cube is you to the 4/4 from pi Over four to I over to and are you is equal to host Sign faded in case you forgot. Now we're just gonna plug that value. And also, this is all times. Um, so we have negative three times co sign before data over four minus Well, not minus her. Um, we have a collection of values. It from I report to you pi over two, and now we're gonna plug in her value. So wait, three kinds coastline for it. I dio over minus coast line fourth, I over for over, for so co sign of pi over two is equal to zero. So this is just all gonna be with cirrhosis is gonna cross out, and so we'll have negative 16/3 from negative. Um, on. So the co sign of pi over four is equal to the square root of to over two. That's going to be for over for on DSO we're gonna have this is equal to 16 over 12 times. Um So the pot square root of to the four is going to be equal to for on our denominator is going to be equal to 16 cause too four is 16 and so we're going to have that this is equal to one third on that is our answer.

You want to find the volume bounded by this parable, Lloyd? Uh, and the planes equal seven, which is the light blue plane up here. But only in the first auction. I have a drawn all the way around. Okay, so first, let's switch to polar. So the, um, problema ideas equals one plus two are square and the plane Z equals seven. Okay, so we're gonna have these pieces so they're gonna be bounded on the top by seven and bounded off the bottom by the parable Lloyd. And then we're gonna look in the first OC tent, and so we're gonna look at what, This circle of intersection down here in the first OC tent. So we're gonna set, um, equal to each other. Seven equals one plus two are square, so six equals to R squared. So R squared is three. So are is the square root of three. Alright, so here's what we have. So we're going to go zero to pi over two zero to the square root of three. Because we're coming out like this. Uh, top minus bottom. Already already. Fatum so is your in a pi over two zero to this word? of 37 minus one that six. My Let's do six are minus two are cute, the R D theta. Okay, so we get zero pi over two six r squared over two. So three r squared minus two are to the fourth over. Four. So are to the fourth over to zero squared of three D theta. So is your the pirate or two square to three square? There's three that's nine minus nine halfs d theta. So that's nine halfs data from zero power to so nine pi over four.

And this question we're going to evaluate a triple integral that we can use either spherical or cylindrical coordinates. To evaluate for this question. I'm going to choose to use sphere. I'm going to choose to use cylindrical coordinates because they're a little bit easier to deal with. So let's remember those conversions now let's remember that. In cylindrical we have X equals R times co sign of data. Yeah. Why equals R times sine of theta? Yeah. NZ is just Z. Of course simply put our region of integration. E is now above this plane. Z equals we know X squared plus y squared. It should be a little bit familiar. Is Bill is above the plane R squared and below the plane Z equals two times are times the sine of phi That's a sign of data. So what does that mean for us? Let's go. So our top part it means that we have this is the top part being below two times are times sine of data. Okay, By equating these together, we can find our bounds to. Yeah. So we have our equals that song. Two R equals two times the sine of theta. Right? That means we actually have our our bounds to So we have z bounds we have our bounds and we have fate abounds. And since we're only and we're only going to consider the upper half of the circle. That is That is from 0 to Pi. That is also Data will be in between zero and pi which means are integral over the region. E of our Small in to grant is going to become an integral from 0 to Pi. An integral from right From 0 to 2 times the sine of Theta. And an integral from R squared to two times are times the sine of theta. And we have our just Z multiplied by the differential volume which remember is our times, e times the Z. Uh huh times DZ d r d Theta. And I just used Wolfram Alpha to evaluate this integral because it asks you you can either use a table or you can use a computer algebra system. So I just used Wolfram Alpha. So we get when we do, so we get the answer of this integral to be five pi over six. And henceforth we have found the answer to our question. Okay, if you wanted to do this with a table, you would have to integrate a couple of times and then eventually end up having to integrate signed to the sixth power of some variable T, which is not very fun and does require you to use an integration table.

In this question does question specifies that our region is a sphere, so data would vary from 0 to 2 pi, and the question specifies that five would vary from zero to pi developed by two, and it's for roll. The lower limit is called sci Fi, and the upper limit is one. We can first rewrite our integral as such, and we can separate it into two integral roles. So, um, in multiplication between two into growth, the first with respect to theta the second with respect to roll and with respect and then took with respect to fight so integrating the second term by with respect to roll, we are essentially integrating role square, so that would give us real cute. Divided by three, substituting in the upper limit and lower limit and expanding it, we would have sci fi, so the first time will be sci fi, and the second term would be signed five times. Cool Sci Fi cube We can break break it down into So instead, of course, sci Fi Cube. We have co sign FIDE multiplied by co sign five square. We can then combine sci fi and cool sci fi to form half next half times signed to five and then consigns. Five square can be rewritten as half times one plus consigned to five expanding it. We can then integrate each term directly. So integrating sine five, we get negative co sign five integrating negative sign to five development four we have consigned to five. Developed by eight. Um, don't forget toe integrate the insight. So because we have 25 we need to divide the whole term by two and then to integrate Negative signed to five, consigned to five developed by four. Do know that this is a case off integrating f x Times f prime X In this case, FX is consigned. Um, because I am to five f Prime X is negative two negative sign to five times too. So one quarter becomes 18 and then we simply increase the power off f x by one and divide it by the new power. So fx being consigned to five would become consigned to five square and the new power being too. We, um, multiply toe by half, substituting inthe e upper and lower limits and simplifying. We have this integral to be equals two high developed by two


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