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”, except that we are For the remaining problems, use the coordinate system (Cartesian, cylindrical,
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Triple Integrals in Cylindrical or Spherical Coordinates
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.