Calculus Stewart ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area
1. Here z= f (x,y)=2+3x+4y and D is the rectangle 0,5 surface is A(S) =
D
1,4 , so by Formula 2 the area of the dA
D
[ f (x,y)] +[ f (x,y)] +1 dA=
x y D
2
2
3 +4 +1 dA= 26
2
2
=
26 A(D)= 26 (5)(3)=15 26
2 2
2. z= f (x,y)=10 2x 5y and D is the disk x +y A(S) =
D
9 , so by Formula 2
( 2) +( 5) +1 dA= 30
D
2
2
dA= 30 A(D)
=
30 (
3 )=9 30
2
3. z= f (x,y)=6 3x 2y which intersects the xy plane in the line 3x+2y=6 , so D is the triangular region 3 . Thus given by ( x,y ) 0 x 2,0 y 3 x 2
{
D
}
A(S) = = 14
( 3) +( 2) +1 dA= 14
D
2
2
dA= 14 A(D)
1 2 3 =3 14 2
2
4. z= f (x,y)=1+3x+2y with 0 x 2y , 0 y 1 . Thus by Formula 2, A(S) =
D 1
(3) +(4y) +1 dA=
0 0 2
2
2
1 2y
10+16y dxdy=
0 1 0
2
1
10+16y
2
x
x=2y x=0
dy
= 2y
0 2 2
10+16y dy= 2
1 2 2 3/2 (10+16y ) 32 3
2 1/2
=
1 3/2 3/2 (26 10 ) 24
5. y +z =9
42
z= 9 y . f =0 , f = y(9 y )
x y
2
A(S) =
0 0 42
0 +[ y(9 y )
2
2 1/2 2
42 0 0
] +1 dydx= y 3
y=2 y=0
y
2 2
+1 dydx 2 3
4
9 y dx=3
=
0 0
3 9 y
2
4
dydx=3
0
sin
1
sin
1
x
0
=12sin
1
2 3
1
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area
6. z= f (x,y)=4 x y and D is the projection of the paraboloid z=4 x y onto the xy plane, that is, D= (x,y)| x +y A(S) =
D 2
2
2
2
2
{
2
2
4
2
}
. So f = 2x , f = 2y
x y 2
( 2x) +( 2y) +1 dA=
D
4(x +y )+1 dA=
0 0 2
2
2
2
2
4r +1 r dr d
2
=
0
1 2 3/2 (4r +1) 12
2 2
r=2 r=0
d =
0 2 2
1 ( 17 17 1 ) d = 6 ( 17 17 1 ) 12
7. z= f (x,y)=y x with 1 x +y A(S) =
D
2
4 . Then 1+4r r dr d =
2 2 2
1+4x +4y dA=
0 1 2 0
2
2
2
d
0 1
r
1+4r dr
2
=
1 2 3/2 (1+4r ) 12
2 1
=
6
( 17
17 5 5 )
8. z= f (x,y)= A(S) =
D 1
2 3/2 3/2 1/2 1/2 (x +y ) and D= { (x,y) 0 x 1,0 y 1 } . Then f =x , f =y and x y 3
(
x
) +(
2
y ) +1 dA=
00 y=1
2
11
x+y+1 dydx
1 0 1 0 3/2 3/2
=
0
2 3/2 (x+y+1) 3
2 dx= y=0 3
(x+2) =
(x+1)
dx
2 2 5/2 2 5/2 (x+2) (x+1) 3 5 5 4 5/2 7/2 = (3 2 +1) 15 = 9. z= f (x,y)=xy with 0 x +y A(S) =
D 2 2 2 1
4 5/2 5/2 5/2 (3 2 2 +1) 15
1 , so f =y , f =x
x y
y +x +1 dA=
0 0
2
2
2
r +1 r dr d =
0
2
2
1 2 3/2 (r +1) 3
r=1 r=0
d
=
0
1 2 ( 2 2 1) d = 3 3
(2
2 1)
2
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area
10. Given the sphere x +y +z =4 , when z=1 , we get x +y =3 so D= (x,y)| x +y z= f (x,y)= 4 x y . Thus A(S) =
D 2 3 0 2 2
2
2
2
2
2
{
2
2
3
}
and
[( x)(4 x y ) r
2 2
2
2 1/2 2
] +[( y)(4 x y )
2 3 0
2
2 1/2 2
] +1 dA
=
0 2
+1 r dr d =
0 2
2r 4 r
2
dr d
4 r 2(4 r )
=
0 2
2 1/2 r= 3 r=0
d = ( 2+4)d = 2
0
2 0
=4
11. z= a x y , z = x(a x y )
x
2
2
2
2
2 1/2
, z = y(a x y )
y
2
2
2 1/2
,
A(S) =
D /2 acos
x +y
2 2
2
2 2
+1 dA
2 2
a x y
=
/2 0 /2 acos
r
2
+1 r dr d
a r ar a r a r
2 2 2 2
=
/2 /2 0
dr d
r=acos r=0
=
/2
a
d
/2
=
/2
a
(
a a cos
2
2
2
a d =2a
)
2
/2 0
(1
1 cos
2
)d
3
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area
=2a
2
/2
d
0
2a
2
/2
sin
0
2
d =a
2
2a
2
/2 0
sin d =a (
2
2)
12. To find the region D : z=x +y implies z+z =4z or z 3z=0 . Thus z=0 or z=3 are the planes where the surfaces intersect. But x +y +z =4z implies x +y +(z 2) =4 , so z=3 intersects the upper hemisphere. Thus (z 2) =4 x y or z=2+ 4 x y . Therefore D is the region inside the circle x +y +(3 2) =4 , that is, D= (x,y)| x +y A(S) =
D 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2
2
2
2
2
{
2
2
3
}
.
2 1/2 2
1+[( x)(4 x y ) r
2 2
2
2 1/2 2
] +[( y)(4 x y )
2 3 0
2
] dA
2 2 1/2 r= 3 r=0
=
0 2 0
1+
r dr d =
0 2 0
2r dr 4 r
2
d =
0
2(4 r )
d
4 r
= ( 2+4)d = 2
0
=4
x y
2 2
13. z= f (x,y)=e A(S)=
x +y
2 2
x y
2
2
, f = 2xe
x x y 2
2 2
, f = 2ye
y x y 2
2 2
x y
2
2
. Then 4(x +y )e
2 2 2(x +y )
2 2
( 2xe
4
) +( 2ye
) +1 dA=
x +y
2 2
+1 dA .
4
Converting to polar coordinates we have
2 2
A(S) =
0 0 2
4r e
2
2
2r
2
2
2
+1 r dr d =
0
2
d
0
r
4r e
2
2r
2
+1 dr
=2
0
r
4r e
2r
+1 dr 13.9783 using a calculator.
14. z= f (x,y)=cos (x +y ) , f = 2xsin (x +y ) , f = 2ysin (x +y ) .
x y
2
2
2
2
2
2
A(S) = =
4x sin (x +y )+4y sin (x +y )+1 dA
2 2
2
2 2
2
2
2 2
2
x +y
1
4(x +y )sin (x +y )+1 dA
2 2
2
2
2 2
2
x +y
1
4
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area
Converting polar to coordinates gives
2 1
A(S) =
0 0 1
4r sin (r )+1 r dr d =
0 2 2 2
2
2 2
2
1
d
0
r
4r sin (r )+1 dr
2
2 2
=2
0
r
4r sin (r )+1 dr 4.1073 using a calculator.
15. (a) The midpoints of the four squares are Here f (x,y)=x +y , so the Midpoint Rule gives A(S) =
D 2 2
1 1 , 4 4
2
,
1 3 , 4 4
2
,
3 1 , 4 4
, and
3 3 , 4 4
.
[ f (x,y)] +[ f (x,y)] +1 dA=
x y D 2
2
2
(2x) +(2y) +1 dA
2
1 4 + = 1 4
2 2 3 +2 2
1 4 3 4
2
+ 2 + 2 7 + 2 11 2
1 4 1 4
2
+1 + +1 + 1.8279
2 2
1 4 3 4
2
2
+ 2 + 2
3 4 3 4
2
2
+1 +1
(b) A CAS estimates the integral to be A(S)=
D
1+(2x) +(2y) dA=
0 0
2
2
1 1
1+4x +4y dydx 1.8616 .
2
2
This agrees with the Midpoint estimate only in the first decimal place. 16. (a) With m=n=2 we have four squares with midpoints 3 3 , 2 2 A(S) =
D
1 1 , 2 2
,
1 3 , 2 2
,
3 1 , 2 2
, and
. Since z=xy+x +y , the Midpoint Rule gives [ f (x,y)] +[ f (x,y)] +1 dA=
x y D 2 2
2
2
(y+2x) +(x+2y) +1 dA 5 2
2
2
2
1 =
3 2
2
+
3 2
2
+1 +
+
7 2
2
+1 +
7 2
2
+
5 2
2
+1 +
9 2
2
+
9 2
22 78 78 166 + + + 2 2 2 2
17.619
5
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area
(b) Using a CAS, we have A(S)=
D
(y+2x) +(x+2y) +1 dA=
00
2
2
22
1+(y+2x) +(x+2y) dydx 17.7165 . This is within about 0.1
2
2
of the Midpoint Rule estimate. 17. z=1+2x+3y+4y , so A(S) =
D 41 2
1+
z x
2
+
2
z y
2
41
dA=
1 0
1+4+(3+8y) dydx
2
=
1 0
14+48y+64y dydx .
Using a CAS, we have
41 1 0
14+48y+64y dydx= or 45 8 14 +
2
45 8
14 +
15 15 ln ( 11 5+3 14 5 ) ln ( 3 5+ 14 5 ) 16 16
11 5+3 70 15 ln . 16 3 5+ 70
2
18. f ( x,y ) =1+x+y+x
1 1
f =1+2x , f =1 . We use a CAS to calculate the integral
x y 1 1 2 1
A(S)=
2 1
f + f +1 dydx=
x y 1
2
2
(1+2x) +2 dydx=2
2
2
1
4x +4x+3 dx and find that
2
A(S)=3 11 +2sinh
3 2 2
or A ( S ) =3 11 +ln ( 10+3 11
)
.
19. f (x,y)=1+x y
1 1 x
2
2 2
f =2xy , f =2x y . We use a CAS (with precision reduced to five significant
x y 1 1 1 x
2
2
2
digits, to speed up the calculation) to estimate the integral A(S)=
1 1 x
2
f + f +1 dydx=
x y 1 x
2
2
2
4x y +4x y +1 dydx , and find that A(S) 3.3213 .
2 4
4 2
6
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area
20. Let f (x,y)=
1+x
2
1+y
1 1 |x|
2
. Then f =
x 2 x 2 y
2x 1+y
2
, f = 1+x
y
(
2
)
2y
( 1+y )
2 2
=
2y 1+x
(
2
)
( 1+y )
2 2
. We use a CAS
to estimate
1 ( 1 |x| )
f + f +1 dydx 2.6959 . In order to graph only the part of the surface above y 1 x as the y range in our plot command.
the square, we use ( 1 x )
21. Here z= f (x,y)=ax+by+c , f (x,y)=a , f (x,y)=b , so
x y
A(S)=
D
a +b +1 dA= a +b +1
D
2
2
2
2
dA= a +b +1 A(D) .
2
2
22. Let S be the upper hemisphere. Then z= f (x,y)= a x y , so A(S) =
D
2
2
2
[ x(a x y ) x +y
D 2 2 2 2 2
2
2
2 1/2 2
] +[ y(a x y )
2 t
2
2
2 1/2 2
] +1 dA
=
2 t
+1 dA=
0 0
r
2
2 2
+1 r dr d
t 0
a x y ar a r
2
a r a
=lim
t a 0 0
dr d =2 lim
2 2 t a
a r
2
2
=2 lim a
t a
a t
2 2
a
=2 ( a)( a)=2 a . Thus the surface area of the entire sphere is 4 a . 23. If we project the surface onto the xz plane, then the surface lies ``above'' the disk x +z the xz plane. We have y= f (x,z)=x +z and, adapting Formula 2, the area of the surface is A(S)=
x +z
2 2
2
2
2
25 in
2
2
[ f (x,z)] +[ f (x,z)] +1 dA=
x z 25 x +z
2 2
2
2
4x +4z +1 dA
25
7
2
2
Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.6 Surface Area
Converting to polar coordinates x=rcos
2 5
, z=rsin
2 1/2
we have
2 0
A(S) =
0 0
4r +1 r dr d =
0
2
2
5
d
0
r(4r +1)
dr=
1 2 3/2 (4r +1) 12
5 0
=
6
( 101
101 1 )
24. First we find the area of the face of the surface that intersects the positive y axis. As in Exercise 23 , we can project the face onto the xz plane, so the surface lies ``above'' the disk x +z z= f (x,z)= 1 z and the area is A(S) = [ f (x,z)] +[ f (x,z)] +1 dA=
2 2
2
2
1 . Then
2
2
2
x
z
0+
2 2
z 1 z
2
2
+1 dA
x +z
1 2 2 1 1 z
2
x +z
1
=
x +z 1
2 2
z
1 1 z 0
2
+1 dA=
1 1 z
2
1 1 z
2
dxdz
1 z 1 1 z
=4
0
dxdz [by the symmetry of the surface]
2
This integral is improper (when z=1 ), so
t 1 z 0
2
A ( S ) =lim 4
t 1 0
1
2
t
dxdz=lim 4
0
1 z
2 2
t
dz=lim 4 dz=lim 4t=4 .
t 1 t 1 0 t 1 1 z 1 z Since the complete surface consists of four congruent faces, the total surface area is 4(4)=16 .
8