2015-test-review-002

Review of double and triple integrals .

1.Find the volume of the solid hemispherical ball

D ={(x,y,z )∈R 3|x 2+y 2+z 2≤a 2,z ≥0}

Solution .By projecting the solid D onto xy -plane,the shadow R of D is a

circular disc of radius a centered at (0,0,0)on xy -plane.Hence R ={(x,y,0)|x 2+y 2≤a 2}={(x,y,0)|?a ≤x ≤a,?√a 2?x 2≤y ≤+√

a 2?x 2}.The region D can be described in terms of rectangular coordinates as {(x,y,z )∈R 3

|?a ≤x ≤a,?√a 2?x 2≤y ≤+√a 2?x 2,and 0≤z ≤+√

?x ?y }.

Then the volume of D can be written as

∫∫∫D

1dV =

∫∫R ∫z max (x,y )

z min (x,y )

1dz dA

∫a ?a

∫√a 2?x 2?√

a 2?x 2

∫+√a 222

1dz dy dx.

However,it would be better for calculation that one rewrite D in terms of spheri-cal coordinates as follows:{(ρ,?,θ)∈R 3|0≤ρ≤a,0≤?≤π/2,0≤θ≤2π}.

∫∫∫D 1dV =∫2π0∫π/20∫a

1·ρ2sin ? Jacobian

dρd?dθ.

2.Find the volume of the cap region D bounded above by x 2+y 2+z 2=a 2and

bounded below by the plane z =

Solution .First,we determine the intersection curve C of these two surfaces:

x 2+y 2+z 2=a 2and z =h.For any point (x,y,z )∈C,we have z =h,and a 2=x 2+y 2+h 2,so x 2+y 2=(a 2?h 2).So the curve

C ={(x,y,h )∈R 3|x 2+y 2=(a 2?h 2)}is a circle in space R 3.

Again project the region D onto xy -plane,its shadow

R ={(x,y,0)|x 2+y 2≤a 2?h 2

}in xy -coordinates,and

={(r,θ)|0≤r ≤√a 2?h 2,0≤θ≤2π}in polar coordinates.

Because of the nature region D,we describe it in terms of cylindrical coordinates,

Then

∫∫∫D

1dV =

∫∫R ∫

z max (x,y )=

√

a 2?x 2?y 2

z min (x,y )=h 1dz dA

∫2π0

∫√a 2?h 20

∫√a 2?r 2

r dz dr theta.

3.Find the volume of the region D bounded above by x 2+y 2+z 2=a 2and bounded below by the positive cone z =λ√

x 2+y 2,where λ=h √a 2?h 2

.Solution .If we compare this question with the previous one,we should know that we need just to change the z min (x,y )in the triple integral.With this in mind,the sphere and the cone can be described as r 2+z 2=a 2and z =λr,i.e.a 2=(1+λ2)r 2,and hence we have z =h,and r =√

a 2?h 2,so the curve of intersection of sphere and the positive cone is a circle on the plane C ={(x,y,h )|x 2+y 2=a 2?h 2},then ∫∫∫D

1dV =∫∫R ∫z max (x,y )=√a 2?x 2?y 2

z min (x,y )=λr 1dz dA

∫2π0

∫√a 2?h 20

∫√a 2?r 2

λr

r dz dr dθ.

4.Find the volume of the region D bounded above by z =h and bounded below by the positive cone z =√

x 2+y 2.Solution .Similar to the previous question,we have ∫∫∫D

1dV =

∫2π0

∫√a 2?h 20

∫h

hr √

a 2?h 2

r dz dr dθ.

5.Evaluate the double integral ∫∫

xy dA,where R is the region on xy-plane bound-

ed by the curves y=x2,and x=y3

Solution.One can?rst?nd the intersection point of these curves as follows: y=x2=(y3)2=y6,i.e.0=y?y6=y(1?y5),so y=0or y5=1,i.e.

y=5

√

1=1,hence y=0or y=1.Correspondingly,we have only two intersec-tion points A(0,0),B(1,1)∈C.It follows that

R={(x,y)∈R2|0≤x≤1,x2≤y≤3

√

x}if one projects R onto x-axis, ={(x,y)∈R2|0≤y≤1,y3≤x≤

√

y}if one projects R onto y-axis.

Then the double integral

∫∫

xy dA=

∫1

∫y

max

(x)=3

√

y min(x)=x2

xy dy dx=

,or

∫1

∫x

max

(y)=

√

x min(y)=y3

xy dx dy=

6.De?nition.the average value of a function f(x,y)(or f(x,y,z)respectively)

de?ned in a region R of R2(or R3respectively).

Average[f]=

∫∫

f(x,y)dA

∫∫

1dA

,or=

∫∫∫

f(x,y,z)dV

∫∫∫

1dV