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Find the volume of the region bounded above by the paraboloid $z=9-x^{2}-y^{2},$ below by the $x y$ -plane, and lying outside the cylinder $x^{2}+y^{2}=1$

$32 \pi$

Calculus 3

Chapter 15

Multiple Integrals

Section 7

Triple Integrals in Cylindrical and Spherical Coordinates

Johns Hopkins University

Oregon State University

Harvey Mudd College

Idaho State University

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.

12:22

Find the volume of the reg…

05:31

05:06

06:51

Find the form of the vision that is is bounded above bound by the paraboloid c, is equal to 9 minus x square plus y squared and is bounded below by the plane by the x y plane. This is the plane, c is equal to 0 and that lies outside of out. This is in their x. Squared plus y square. Equals 1 is very exciting. So out of this technically have this very nice region. We have here a paraboloid minus, so you call this 1. The c axis the x and y of these should be the barbaroi 9 minus x. Squared minus y square is equal to z. Then we have the plane here. So it is this region vice outside of these encilinders. You have here the cylinder where we move that i will obtain some region like this also first well, you can see that this is symmetrical with respect to that, because this is symmetric with respect to the in cylindrical coordinates that will be very convenient. Writing cylindrical, because its symmetric with respect to the angle symmetrical coordinates. So while this condition is saying there are are square smaller than 1, and these all these can be translated. This condition say that is our square, so that, for this volume or c will be between 0 and 9 minus r, squared the as you can see here. If we draw the r and they c you have here the curve on 99 minus r square equals to c so our r riveting 1 and so retries c times of r, then we would need to find out what is this point not so that at that Point you have that t is equal to 0, so we're going to have that 9 minus r, not square, is equal to 0, so that or not square is equal to 9. Moving that to the other side, and so whethase r is positive or not is 3. Is positive so so that our radius should go from 0, not from 1. You see this outside of this goes from argo, spin, 1 up 23 and then by the symmetry of theta goes all the way around theta goes from 0 up to 2 pi. So for this volume you going to have all this should be the vands for c from 0 up to 9 minus r square that should be with the volyume element is intgal. So then we have the bonds for r between 1 and 3 are they are from 0 to pi 0 to pi, so the volume is gottan so well. We can integrate keeping the order of derision in 12, o c j talate within 9 minus r, squared and 0 point. This is the 9 r squared minus 0 that is 0, so we're going to have 1. I draw from 1 up to 3 of 9 minus r, squared times r r you have this integral, and that goes from 0 up to 2 pi to pi. So here we have 2 integrals and draw 9 r on intialo r cube. So this velamine r is by the power of 9 r square halves, and the total r c would be r to the fourth divity 4 divided by 4. So it would be the valaitis between 1 and 3. Piero here is v 1, so you have that this is going to be equal to 9 times 3 squared half 3 to the 4 power divided by 4 minus so 9 halves times 1 squared minus 14. Also here that is that so here our 9 is equal to 3 square, so that this whole term will be 3 to the 4, so halves minus 3 to the orandi 4, and so all this can be seen as 3 to the 4 power times. 1. Half minus 1- and this is equal to its equal to 2, so we'll be equal to just 1 for that substruction order is 1 e, 4 times 3 to the fourth power, and this term here we have is the same as 18 multiplying by 2 in the Bottom, so would have 184 misthis should be equal to 17 over 4, so this will be equal to minus 10 o 4. To this we had to be to forecast 9 times 9 minus 17, o 49 times 1 minus 17. So this would be equal to 6. Then 66, so this will be equal to 66 potatie to since this 66 is equal to 33 times 2. That would cancel 12 there so have a very 3 halves so that this this integral is far 3 halves, and so like these, you turn into a drop from 0 up to 2 pi 33 halesand. So all this interval have visus theta evaluated between 2 pi and 0 point, so this would be 2 pi, sothades moingo be 2 pi times there 3 halves, and so all these councels and the volume will be equal to 3. So that's going to be the volume of the vision or for that region or will be equal to 33, a pi.

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