This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 0 10 − 2x
5 dx 11. Let Q be the region between z = (x2 + y 2 )3/2 and z = 1, and inside x2 + y 2 = 4. Sketch the region Q, and then
x2 + y 2 ez dV as an integral in the best(for this example) 3dimensional coordinate system. DO NOT write
Q EVALUATE THE INTEGRAL.
z 8
z = r3 y
1
1 x
2 Given the form of the solid region and the function, cylindrical coordiates is the best system to use to express this
integral.
Converting the top and bottom surfaces, we have z = r2 3
2 = r3 and z = 1 Notice that the intersection between these two surfaces is: (x2 + y 2 )3/2 = 1, or x2 + y 2 = 1, the circle of radius 1 about
the origin.
The region in the plane is all points between the circle of radius one and the circle of radius 2 both centered at the
origin, and the integrand becomes rez .
r3 2 2π r3 2 2π r2 ez dz dr dθ (rez ) r dz dr dθ = Thus the integral is:
0 1 0 1 1 1 12. Set up a triple integral in spherical coordinates for the volume V of the region between
z = 3x2 + 3y 2 and the sphere x2 + y 2 + z 2 = 16. Be sure to include a sketch of the region,
and DO NOT EVALUATE THE INTEGRAL.
z 4 ρ=4
3 y x 3x2 + 3y 2 , we have z 2 = 3x2 + 3y 2 , or 4z 2 = 3x2 + 3y 2 + 3z 2 = 3ρ2 Translating z =
2 Then 4 (ρ cos φ) = 3ρ2 , or 4ρ2 cos2 φ = 3ρ2 . Thus we have 4 cos2 φ = 3, or cos2 φ = 3 .
4
Hence cos φ = ± √ 3
2.
2 Since we only want values between 0 and π , we take φ = π
6, [a 30◦ cone]. Notice that x2 + y + z 2 = 16 is the sphere of readius 4 about the origin, and has the form ρ = 4.
2π 4 π
6 ρ2 sin φ dρ dφ dθ Therefore, the integral has the form
0 0 0 3x2 + 3y 2 and z = 13. (a) Let Q be the region between z = 4 − x2 − y 2 . Sketch the region Q z 2 ρ=2
3 y x (b) Set up, but do not evaluate a triple integral in rectangular coordinates that gives the volume of Q.
√ 22
√
2
1− x 1 1 dV = 4 Using symmetry, 4− x − y 1 dz dy dx 3x2 +3y 2 0 0 Q √ (c) Set up, but do not evaluate a triple integral in cylindrical coordinates that gives the volume of Q.
1 π
2 Again using symmetry, 1 dV = 4
Q √ 0 0 √ 4− r 2 1 r dz dr dθ 3r (d) Set up, but do not evaluate a triple integral in spherical coordinates that gives the volume of Q.
We need to ﬁgure out the angle of declination for the cone. Starting with z 2 = 3x2 + 3y 2 , we see that 4z 2 =
3x2 + 3y 2 + 3z 2 , or 4z 2 = 3ρ2 .
3
Therefore, since z = ρ cos φ, 4ρ2 cos2 φ = 3ρ2 . Hence 4 cos2 φ = 3, or cos2 φ = .
4
√
3
π
. The solution to this in the ﬁrst quadrant is φ = .
Thus cos φ = ±
2
6
2π 2 π
6 ρ2 sin φ dρ dφ dθ 1 dV = So we have 0 0 Q 0 (e) Pick one of the triple integrals you found above and evaluate it in order to ﬁnd the volume of Q exactly.
2π ρ sin φ dρ dφ dθ =
0 π
6 =
0 8
3 0 8
8
sin φ dφ dθ =
3
3
π
6 2π
0 − cos φ 0 8
dθ =
3 0 0 0 0 0 13
ρ sin φ
3 2 dφ dθ
0 π
6 2π
0 − cos φ − √
3
−1
2 2π π
6 2 We will use the spherical integral:
2π 2π 2 π
6 dθ
0 8
dθ =
3 √
1− 3
2 2π θ
0 √
8
3
= (2π ) 1 −
3
2 16π
=
3 √
3
1−
2...
View Full
Document
 Fall '11
 James
 Math, Vector Calculus

Click to edit the document details