Triple Integrals in Cylindrical and Spherical Coordinates
Note: Remember that in polar coordinates dA = r dr d . EX 1 Find the volume of the solid bounded above by the sphere x2 + y2 + z2 = 9 below by the plane z
Review for Exam 3. Triple integral in spherical coordinates (Sect
2. ) hence A = 3π + 9. √. 3/2. <. Page 6. Double integrals in Cartesian coordinates. (Sect. 15.2). Example. Find the y-component of the centroid vector in
MATH 20550 Triple Integrals in cylindrical and spherical coordinates
into a spherical coordinate iterated integral. (from here example 2.) Let us start by describing the solid. Note ∫. 3. 0. ∫. √.
Triple Integrals in Cylindrical Coordinates Many applications involve
In particular there are many applications in which the use of triple integrals is more natural in either cylindrical or spherical coordinates. For example
15.8: Triple Integrals in Spherical Coordinates
We have: Page 6. Example. Let's get a better handle on things by graphing some basic functions given in spherical coordinates. Let c be a constant. Sketch the
April 8: Triple Integrals via Spherical and Cylindrical Coordinates
Apr 8 2020 Example 1. Let's begin as we did with polar coordinates. We want a. 3-dimensional analogue of integrating over a circle. So we integrate ...
Triple Integrals in Spherical Coordinates
For example the sphere with center the origin and radius c has the simple equation ρ = c (see Figure 2); this is the reason for the name “spherical”
Triple Integrals in Cylindrical and Spherical Coordinates
Triple Integrals in Cylindrical and Spherical Coordinates. 29/67. Page 30. How to Integrate in Spherical Coordinates - An Example. Example 5. Find the volume of
3.6 Integration with Cylindrical and Spherical Coordinates
In this section we describe
TRIPLE INTEGRALS IN SPHERICAL COORDINATES EXAMPLE A
TRIPLE INTEGRALS IN SPHERICAL COORDINATES. EXAMPLE A Find an equation in spherical coordinates for the hyperboloid of two sheets with equation . SOLUTION
Review for Exam 3. Triple integral in spherical coordinates (Sect
Line integrals in space. Example. Evaluate the line integral of the function f (xy
Page 1 Section 15.8: Triple Integrals in Spherical Coordinates
Spherical Coordinates: A Cartesian point (x y
15.8: Triple Integrals in Spherical Coordinates
We have: Page 6. Example. Let's get a better handle on things by graphing some basic functions given in spherical coordinates. Let c be a constant. Sketch the
MATH 20550 Triple Integrals in cylindrical and spherical coordinates
into a spherical coordinate iterated integral. (from here example 2.) Let us start by describing the solid. Note ?. 3. 0. ?. ?.
Integrals in cylindrical spherical coordinates (Sect. 15.7) Cylindrical
Notice the extra factor ?2 sin(?) on the right-hand side. Triple integral in spherical coordinates. Example. Find the volume of a sphere of radius R.
Triple Integrals in Cylindrical Coordinates Many applications involve
In particular there are many applications in which the use of triple integrals is more natural in either cylindrical or spherical coordinates. For example
Triple Integrals in Spherical Coordinates
For example the sphere with center the origin and radius c has the simple equation ? = c (see Figure 2); this is the reason for the name “spherical”
Triple Integrals in Cylindrical or Spherical Coordinates
xyz dV as an iterated integral in cylindrical coordinates. x y z. Solution. This is the same problem as #3 on the worksheet “Triple Integrals”
15.7 Triple Integrals in Cylindrical and Spherical Coordinates
Figure 15.44 Page 894. Example. Page 901
Section 9.7/12.8: Triple Integrals in Cylindrical and Spherical
from rectangular to spherical coordinates. Solution: ·. Example 7: Convert the equation ? ? sec2. =.
1.LetUbe the solid enclosed by the paraboloidsz=x2+y2andz= 8(x2+y2). (Note: The paraboloids
intersect wherez= 4.) WriteZZZ Uxyz 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 r2(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 r2(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 U1dV. 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
U1dVas an iterated integral in spherical
1 coordinates Z 2 0Z 0Z 1 012sin d d d=
Z 2 0Z 0 33sin=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 U1dV. 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 012sin 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 U1dV. 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 zyWe'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 01r dr d.
Putting this into our outer integral, we get the iterated integralZ 2 1Z 2 0Z z 01r 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 withdzasthe 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=px2+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. 4Let'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 r2rcosr dz dr d.
7.The iterated integral in spherical coordinatesZ
=2Z =2 0Z 2 13sin3 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_dbs21.pdfusesText_27[PDF] trivia questions pdf
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