I7-26. 2 π
I7-27.
19
24I7-28. π
I7-30. If~r=~r(u,v), thend~r=∂~r∂udu+∂~r∂vdvand
ds^2 =d~r·d~r=∂~r
∂u·∂~r
∂u(du)(^2) + 2∂~r
∂u·
∂~r
∂vdudv+
∂~r
∂v·
∂~r
∂v(dv)
2
I7-31. (b) S=πR
√
R^2 +H^2
I7-32.
dS =ρ(a+ρcosθ)dθdφand dV =ρ(a+ρcosθ)dρdθdφ Volume is V = 2π^2 aρ^2 and
surface area isS= 4π^2 aρ
What does the Pappus theorem tell you about the volume?
I7-33. (d)ds=
√
1 + (y′)^2 dxwherey=x^2 , y′= 2xso that
S=
∫ 2
0
√
1 + (2x)^2 dx
To evaluate this integral make the substitution 2 x= sinhuwith 2 dx= coshuduand
show
S=
∫ sinh−^1 (4)
0
1
2
cosh^2 udu=^1
2
∫ sinh−^1 (4)
0
[^1
2
cosh2u+^1
2
]du
Show that
S=
√
17 +
1
4 sinh
− (^1) (4)
Solutions Chapter 7