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ated integral in polar coordinates to describe this disk: the disk is 0 ≤ r ≤ 2, 0 ≤ θ < 2π, so our iterated To compute this, we need to convert the triple integral an iterated integral which gives the volume of U (You need not evaluate )

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15 4 Double Integrals in Polar Coordinates 15 9 Triple Integrals in Spherical Coordinates Evaluate the integral by changing to cylindrical coordinates: ∫ 1

[PDF] Integrals in cylindrical, spherical coordinates - MSU Math

Triple integral in spherical coordinates Cylindrical coordinates in space Definition The cylindrical coordinates of a point P ∈ R3 is the ordered triple (r, θ, z)

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Evaluate the integral using spherical coordinates: dxdydz T ∫∫∫ 16 10 The Jacobian; Changing Variables in Multiple Integration So far, we have used

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Thus to evaluate an integral in spherical coordinates, we do the follow- ing: (i) Convert the function f(x, y, z) into a spherical function (ii) Change the limits of the

[PDF] Section 1310: Triple Integrals in Cylindrical and Spherical

Some regions in space are easier to express in terms of cylindrical or spherical coordinates Triple integrals over these regions are easier to evaluate by

[PDF] TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL

A Review of Double Integrals in Polar Coordinates where we write ∆r = b−a and ∆θ = d−c (the change in radius and the change (3) Evaluate the integral

[PDF] Triple Integrals in Cylindrical or Spherical Coordinates

[PDF] Homework 24: Cylindrical/Spherical integration

)dV , where H is the solid hemisphere x 2 + y 2 + z 2 ≤ 16, z ≥ 0 4 Evaluate the integral by changing to spherical coordinates ∫ a −a ∫ √ a2−y2 − √

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Unmarked “Homework 10” Solutions 2016 April 8 1 Evaluate ∫ 3 -3 ∫ √ by changing to spherical coordinates Solution: Therefore the given integral is

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(17 points): Evaluate the integral by changing to spherical coordinates ∫ 4 0 ∫ √ 16−y2 − √ 16−y2 ∫ √ 16−x2−y2 0 (x2 + y2 + z2)z dz dx dy

1.LetUbe the solid enclosed by the paraboloidsz=x2+y2andz= 8(x2+y2). (Note: The paraboloids

intersect wherez= 4.) WriteZZZ U

xyz dVas an iterated integral in cylindrical coordinates.xyzSolution.This is the same problem as #3 on the worksheet \Triple Integrals", except that we are

now given a specic integrand. It makes sense to do the problem in cylindrical coordinates since the solid is symmetric about thez-axis. In cylindrical coordinates, the two paraboloids have equations z=r2andz= 8r2. In addition, the integrandxyzis equal to (rcos)(rsin)z.

Let's write the inner integral rst. If we imagine sticking vertical lines through the solid, we can see

that, along any vertical line,zgoes from the bottom paraboloidz=r2to the top paraboloidz= 8r2.

So, our inner integral will beZ

8r2 r

2(rcos)(rsin)z dz.

To write the outer two integrals, we want to describe the projection of the solid onto thexy-plane. As we had gured out last time, the projection was the diskx2+y24. We can write an iter- ated integral in polar coordinates to describe this disk: the disk is 0r2, 0 <2, so our iterated integral will just beZ 2 0Z 2 0 (inner integral)r dr d. Therefore, our nal answer isZ 2 0Z 2 0Z 8r2 r

2(rcos)(rsin)zr dz dr d.

2.Find the volume of the solid ballx2+y2+z21.

Solution.LetUbe the ball. We know by #1(a) of the worksheet \Triple Integrals" that the volume ofUis given by the triple integralZZZ U

1dV. To compute this, we need to convert the triple integral

to an iterated integral. The given ball can be described easily in spherical coordinates by the inequalities 01, 0,

0 <2, so we can rewrite the triple integralZZZ

1dVas an iterated integral in spherical

1 coordinates Z 2 0Z 0Z 1 0

12sin d d d=

Z 2 0Z 0 33
sin=1 =0! d d Z 2 0Z 013 sin d d Z 2 0 13 cos= =0! d Z 2 023
d =4 3

3.LetUbe the solid inside both the conez=px

2+y2and the spherex2+y2+z2= 1. Write the triple

integralZZZ U z dVas an iterated integral in spherical coordinates.

Solution.Here is a picture of the solid:xyzWe have to write both the integrand (z) and the solid of integration in spherical coordinates. We know

thatzin Cartesian coordinates is the same ascosin spherical coordinates, so the function we're integrating iscos.

The conez=px

2+y2is the same as=4

in spherical coordinates.(1)The spherex2+y2+z2= 1 is = 1 in spherical coordinates. So, the solid can be described in spherical coordinates as 01, 0 4 , 02. This means that the iterated integral isZ 2 0Z =4 0Z 1 0 (cos)2sin d d d.

For the remaining problems, use the coordinate system (Cartesian, cylindrical, or spherical) that seems

easiest.

4.LetUbe the \ice cream cone" bounded below byz=p3(x2+y2)and above byx2+y2+z2= 4. Write

an iterated integral which gives the volume ofU. (You need not evaluate.)(1)

Why? We could rst rewritez=px

2+y2in cylindrical coordinates: it'sz=r. In terms of spherical coordinates, this

says thatcos=sin, so cos= sin. That's the same as saying that tan= 1, or=4 2 xyzSolution.We know by #1(a) of the worksheet \Triple Integrals" that the volume ofUis given by the triple integralZZZ U

1dV. The solidUhas a simple description in spherical coordinates, so we will use

spherical coordinates to rewrite the triple integral as an iterated integral. The spherex2+y2+z2= 4 is the same as= 2. The conez=p3(x2+y2) can be written as=6 .(2)So, the volume isZ 2 0Z =6 0Z 2 0

12sin d d d.

5.Write an iterated integral which gives the volume of the solid enclosed byz2=x2+y2,z= 1, and

z= 2. (You need not evaluate.)xyzSolution.We know by #1(a) of the worksheet \Triple Integrals" that the volume ofUis given by

the triple integralZZZ U

1dV. To compute this, we need to convert the triple integral to an iterated

integral. Since the solid is symmetric about thez-axis but doesn't seem to have a simple description in terms of spherical coordinates, we'll use cylindrical coordinates.

Let's think of slicing the solid, using slices parallel to thexy-plane. This means we'll write the outer

integral rst. We're slicing [1;2] on thez-axis, so our outer integral will beZ 2 1 somethingdz.

To write the inner double integral, we want to describe each slice (and, within a slice, we can think of

zas being a constant). Each slice is just the disk enclosed by the circlex2+y2=z2, which is a circle of radiusz:(2)

This is true becausez=p3(x2+y2) can be written in cylindrical coordinates asz=rp3. In terms of spherical coordinates,

this says thatcos=p3sin. That's the same as saying tan=1p3 , or=6 3 ?zzx ?z z

yWe'll use polar coordinates to write the iterated (double) integral describing this slice. The circle can

be described as 0 <2and 0rz(and remember that we are still thinking ofzas a constant), so the appropriate integral isZ 2 0Z z 0

1r dr d.

Putting this into our outer integral, we get the iterated integralZ 2 1Z 2 0Z z 0

1r dr d dz.

Note:For this problem, writing the inner integral rst doesn't work as well, at least not if we want to

write the integral withdzas the inner integral. Why? Well, if we try to write the integral withdzas

the inner integral, we imagine sticking vertical lines through the solid. The problem is that there are

dierent \types" of vertical lines. For instance, along the red line in the picture below,zgoes from the

cone (z=px

2+y2orz=r) toz= 2 (in the solid). But, along the blue line,zgoes fromz= 1 to

z= 2. So, we'd have to write two separate integrals to deal with these two dierent situations. xyz6.LetUbe the solid enclosed byz=x2+y2andz= 9. Rewrite the triple integralZZZ U x dVas an iterated integral. (You need not evaluate, but can you guess what the answer is?) Solution.z=x2+y2describes a paraboloid, so the solid looks like this:xyz Since the solid is symmetric about thez-axis, a good guess is that cylindrical coordinates will make things easier. In cylindrical coordinates, the integrandxis equal torcos. 4

Let's think of slicing the solid, which means we'll write our outer integral rst. If we slice parallel to the

xy-plane, then we are slicing the interval [0;9] on thez-axis, so our outer integral isZ 9 0 somethingdz.

We use the inner two integrals to describe a typical slice; within a slice,zis constant. Each slice is a disk

enclosed by the circlex2+y2=z(which has radiuspz). We know that we can describe this in polar coordinates as 0rpz, 0 <2. So, the inner two integrals will beZ 2 0Z pz 0 (rcos)r dr d. Therefore, the given triple integral is equal to the iterated integralZ 9 0Z 2 0Z pz 0 rcosr dr d dz= Z 9 0Z 2 0 13 r3cosr=pz r=0! dr d dz Z 9 0Z 2 013 z3=2cos d dz Z 9 0 13 z3=2sin=2 =0! dz =0 That the answer is 0 should not be surprising because the integrandf(x;y;z) =xis anti-symmetric about the planex= 0 (this is sort of like saying the function is odd:f(x;y;z) =f(x;y;z)), but the solid is symmetric about the planex= 0.

Note:If you decided to do the inner integral rst, you probably ended up withdzas your inner integral.

In this case, a valid iterated integral isZ

2 0Z 3 0Z 9 r

2rcosr dz dr d.

7.The iterated integral in spherical coordinatesZ

=2Z =2 0Z 2 1

3sin3 d d dcomputes the mass of a

solid. Describe the solid (its shape and its density at any point). Solution.The shape of the solid is described by the region of integration. We can read this o from the bounds of integration: it is 2 , 02 , 12. We can visualize 12 by imagining a solid ball of radius 2 with a solid ball of radius 1 taken out of the middle. 02 tells us we'll only have the top half of that, and 2 tells us that we'll only be looking at one octant: the one withxnegative andypositive:xyz To gure out the density, remember that we get mass by integrating the density. If we call this solid U, then the iterated integral in the problem is the same as the triple integralZZZ U sin2 dVsince dVis2sin d d d. So, the density of the solid at a point (;;) issin2. 5quotesdbs_dbs17.pdfusesText_23

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