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Find the volume of the smaller region cut from the solid sphere $\rho \leq 2$ by the plane $z=1$.

$\frac{5 \pi}{3}$

Calculus 3

Chapter 15

Multiple Integrals

Section 7

Triple Integrals in Cylindrical and Spherical Coordinates

Missouri State University

University of Michigan - Ann Arbor

Idaho State University

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.

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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.

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Use spherical coordinates …

We have at this upper hemisphere radiosone to the x y z axis. We have the mister ashe radius of 2. Resistance is 2 and we make a cut at c equals 1 for a plane for that pantagias going to give us. This is the plan c equals 1. So we want to compute what is the volume of this region in the so ebe the region inside of a sphere, solid sphere, so row listen to and above this plane c equals 1. So how do we do that? To go first, if you have here this oxyton, this is 2 and there is 1 up here x. Axis y equals 0 point. So we have this sphere. Something like that. So the angle not which is happens, is you can figure it out by her is join. That should be like that half tenedivided by 3. This is like 3 pieces, so o this angle, there is equals to 5 third, so that gives us this point all so that fee should be between 0 only 5 third and to the whoever there, because it's symmetric with respect being there. So there goes from 0 up to 2 pi and then our radius row. So, as you can see here here about, the axis goes from 1 up to 2 on, but as you keep on stretching there, the range over a increases so now, for how so the plane c goes, 1 is given by this is grow. Cosine of t equals 1 point, so you have to be above that plane, cosine thee, so row has to be bigger than 1 over cosine of t, and so so you can check that and well. You can check by his looking that at 0 is 1 half sae between 0 and 2 and then at this point by third we're going to have 5 thirds all the 1 over. It is because 1 over cosine of 5 third is equal to 1. Over 1 half, which is equal to 2 point, so you would be saying with this equation, that row has to unfit 2 and 2 if he is, there is no space so that that is the bond for row and that this is a scant of. So this, for comparing this volume you can use spherical coordinates so that we have here was between secant of f and 2. The is the nubra square sine of t, then p goes between 0 and pi third 0 by third and f, and there goes all the way around 0 to 2 pi so that when leaving to get first row square, how there is that is equal to row. Cube! Third, so that these would be equal to that evaluated between 2 and second 2 and c equal to 1. Third, half to 2 cube minus c gant of the cube, and then that's it so be. If meeting now, but have to multiply by sine sine and then between 0 and the third and there, between 0 and 2 pi, so when we have the integral integral sign the first integral there, timsaconstant is going to be equal to minus cosine. And so at that evaluated within pi third and 0 wobanminus cosine of pi third minus cosine of 0 point. So of this number. There is 1 half minus 1 into this minus 1 half with the minus. This is equal to 1, half 1 half positive. So that is this integral so that that would be 1 third it from 0 to pi, and then that is 2 to the third times 1 half and we have lacentleftso these be fair and then minus 0 to 2 pi and then trefrom 0. Parenthesis here to pi third of a sine of p d: it secant is equal to livercosine, so that will be divided by cosine cube. So here we can draw new substitution because you can see that cosine of t is equal to? U? D? U will be equal to minus sine okay, so so that this integral would be d? U over? U square! O! U cube the center! A with the minus sine satin trouble minus sine of t d c, where cosine of the cube will be equal to these! With? U equals to cos sin, and so, if we interets that is equal to minus litre, a on usually squared, so you have a negative power, so minus 1 over 2- u square you differentiate this and get so. It would be that evaluated between article sine. So it would be minus 1 over 2 cosine of t square so that that blow is within i third 0 tistefote 0 pine will give us minus minus 1 half 1 over cosine of my third is 1 half squared otheninus cosine of 0 is 1, so these Would be just 2 squared minus 1 point, so this number is 3 somnusto, be minus 3 halves so that interallie minus 3 halves the centegral over here. This integral is with the minuses minus 3 halves, so that are ground. Putting putting everything together. So our thorough integral is 1 third metre from 0 to pit, so this term will be 2 squared seethe, 2 squared, and all this time was was that minus 3 halves minus 3 halves there. So so this should be our volume and then we do of the theta from got to i pistoia and then what this number there. Its tie up. 4 minus 3 halves with 4, is equal to 8 halves. Minus 3 halves. That will be 5 halves. So this will be times 5 halves. Change to this volume is 5 by third. Soon of this region is 5 by third ball. Riht.

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