Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

(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 hemisphere $\rho=1, z \geq 0$, and above by the cardioid of revolution $\rho=1+\cos \phi$

$11 \pi / 6$

Calculus 3

Chapter 15

Multiple Integrals

Section 7

Triple Integrals in Cylindrical and Spherical Coordinates

Campbell University

Baylor University

University of Nottingham

Boston College

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.

14:10

(a) find the spherical coo…

13:31

05:18

12:26

09:54

11:45

13:05

In Exercises $55-60,$ (a) …

All right, we've got way. Wanna find the circle corner limits? Because we want to be able to calculate. Mm hmm. Volume. So we're gonna use a standard, triple integral problem to integrate with respect to x, Y and Z. And then that way we'll be able to determine the volume of solid. And we can look at the graph to help us out there. And we can see that Ray M. It's gonna enter P equals one, and it will leave at p equals one plus cosign fee. Eso. Then when he was our limit or p the b from one to instead of this PC and mhm one plus co sign feed. Okay, we know fee is gonna vary from zero Pike was too, you know, two pi over two. Hm. So for these are limits. And then the last summit we need to calculate for will be for our theater. And we know that the ray l is going to shoot over our because they don't run from zero two pi The two times power. So we would say, Well, data a equal to zero. Mhm. They, uh mm. Equal to two. Fine. And now in this case, we're gonna have not x y and z, um, putting these in polar coordinates. Um, so we'll say is DP de fee and deep data. Yeah, and keep saying P, but, um, Greek symbol here is row. That's my fault there. But just remember, that's row B data. Okay? These are gonna be your limits. The intervals. We're using polar coordinates, like I said, and we're gonna use a triple integral salt for the volume. And it will be, um, you know, our fi. It'll be So I row, then we integrate with respect to be, and then we respect integrate with respect that data. So for avoiding Okay, now, our equation that we use in the middle is gonna be row squared, sign the we're solving the volume, and the limits will go in the order with fee being here. I'm sorry. Nothing row being there. And you can see that our role limits or from one, 21 plus cosign fee and then the next limit. Yeah, well, b c and we know that zero pi over two. And then the last one is data which we know instruments. Zero to park. Okay. When we solve for this integral. We will get our volume. We will find our final answer. All right, so let's go step by step. So we integrate p squared. We know that he cute over three. Keep saying, P I'm sorry. Row Q over three. And that is integrated one to one plus co sign. Take fee and then we're gonna multiply by sign feet and we'll have it integrated by sign feed Inside Data. We still have our two integral from zero to y 02 then zero. Too far. Okay, Now, um, from here we can expand this out. When we go through and solve forward, we're gonna end up simplifying this and we pull out a one third. The half still are to integral. They're 0 to 2 pi and zero the pile of to. But then inside we have one waas co sign you see and three co sign. Plus the new coach. I squared. He loss. We co signed the waas. I'm sorry. Minus one. Okay. When we break down this integral, that's what we're here. And we still have our sign our d c and D data. Yeah. Alright then pull this down. Now we're gonna integrate this with respect to feed, we're still gonna have a fee on the outside. You have an integral from 0 to 2 pi. Then we integrate into respect to feed. It will take us to every three pie, too. Cosine squared. Mm. Waas start minus close in Cuba. He And then if we integrate that, we will have a negative co sign. What? Hmm? Over four. Can't. We're integrating that with the limit. Zero to pi over two. And we have Delta the students. Okay, I'm sorry. D data will stay. All right. And then when we go through and solve this integral and we plug in the data values of Pi over two and zero, we will end up getting so you want a two pi. And what will happen is we'll get a bunch of zeros here. If I were to cosign that priority, we know zero. And then we're subtracting by born fourth, minus green. What's right? Three thirds when we plug in the zero. And then last one would yield us three over to on who out a negative. Actually, this negative here. Who out there so actually gonna have the negative one third out here? do you think? Okay. And then what we'll end up is we can pull both the negatives out, and we'll end up with one third, and then we integrate this. He's with respect to theta. We'll get add almost together. That's three. Uh, so if we leave, we add all these values together we would have 1/4 plus one. So 1 1/4 was three house way would have Thio use 12 other denominator. It's 33 I was 12 minus 12 on, then minus 18. Yes. Okay, so that will come out. Three plus 12 plus 18 is 33. Okay. When you add all these together right on, they'll still be negative. Essentially negative. But we had a negative out here and we end up getting positive. We pulled the negative out on, and then we have integral zero to pi. You know, on what's gonna end up happening is we're gonna have 33/36 and we'll have a data integrate from 0 to 2 pi, and that will be 33/36 multiplied by two pi minus zero. So we will get a 66 over 36. All right. Uh, huh? Believe I did that correctly. 36 66 or 36 right? Yes, that is correct. Okay, now what? We can dio and everything here should be able to divide by two. So that turned into 33. Bottom one. They're divided by 2 18. Divide both sides by three. 11/6 is what simplifies. Okay, if you divide both sides by six. If I both the top in the bottom here about six. We got six in the denominator six in the numerator. You get 11/6 pie as your final volume there. All right, well, I hope that clarifies. The question is you can see the way we solve for this is just a standard triple and of a problem for solving the volume surface. Um, I'm sorry. Volume of a solid And what we did first is we saw for the limits and we plugged it in into our triple integral problem. We were able to find our final answer for volume, which is pirates six 11/6 times pi. All right, well, I hope that clarifies the question. Thank you for watching

View More Answers From This Book

Find Another Textbook

01:04

Rewrite the equation in the form $a x^{2}+b x+c=0,$ with $a > 0,$ and the…

02:24

Solve for the indicated variable.$$\text { For } x: 5 x-2 y=19$$

01:05

Find the unknown.$$y^{2}+49=0$$

00:53

Find the unknown.$$3 z^{2}-24=0$$

04:12

Solve for the unknown, and then check your solution.$$7 m-4=3 m+12$$

01:42

01:01

06:11

Solve for the unknown, and then check your solution.$$0.35+0.24 x=0.2(5-…

02:19

Complete the square in each of the following by putting in the form $(x+B)^{…

00:56

Find the unknown.$$x^{2}+4=0$$