You should be able to solve any homework/tutorial problem. The exam problems will be very similar to them. The formulae below will be appended to your exam paper. If you need other formulae to solve the problems you should memorize them (or know how to derive them).
Useful Formulae
1. Let r ( t ) be a vector function with values in
R
3 : r ( t ) = x ( t ) i + y ( t ) j + z ( t ) k .
(a) The unit normal vector is N ( t ) = d T
| dt d T dt
|
.
(b) The unit binormal vector is B ( t ) = T ( t ) × N ( t ) .
(c) The curvature of C is κ ( t ) =
| T
0
( t ) |
| r
0
( t ) |
.
2. Let σ be a surface in
R
3 : z = f ( x, y )
(a) The slope k x of the surface in the x -direction at the point ( x
0
, y
0
) is k x
=
(b) The slope k y of the surface in the y -direction at the point ( x
0
, y
0
) is k y
=
∂z
∂x
( x
0
, y
0
) .
∂z
∂y
( x
0
, y
0
) .
(c) The equation for the tangent plane to the surface at the point P = ( x
0
, y
0
, z
0
) is z = z
0
+ k x
( x − x
0
) + k y
( y − y
0
) .
(d) Parametric equations for the normal line to the surface at P = ( x
0
, y
0
, z
0
) are r ( t ) = r
0
+ t ( − k x i − k y j + k ) , r
0
= x
0 i + y
0 j + z
0 k .
(e) The mass of the lamina with the density δ ( x, y, z ) that is the portion of the surface that is above a region R in the xy -plane is r
M =
RR
σ
δ ( x, y, z ) dS =
RR
R
δ ( x, y, z ) 1 +
∂z
∂x
2
+
∂z
∂y
2 dA .
3. Let R be a plain lamina with density δ ( x, y ).
(a) Its mass is equal to M =
RR
R
δ ( x, y ) dA .
(b) The x -coordinate of its centre of gravity is equal to x cg
(c) The y -coordinate of its centre of gravity is equal to y cg
=
=
1
M
1
M
RR
RR
R
R x δ ( x, y ) dA .
y δ ( x, y ) dA .
4. Let G be a solid enclosed between the two surfaces (in spherical coordinates) r = g ( θ , φ ) , r = h ( θ , φ ). The triple integral over the solid is
RRR
G f ( r, θ, φ ) dV =
R
2 π
0
R
π h
R h ( θ ,φ ) f ( r, θ, φ ) r
2 dr i sin φ dφ dθ .
0 g ( θ ,φ )
5. Let a region R xy in the xy -plane be mapped to a region change of variables u = u ( x, y ) , v = v ( x, y ).
R uv in the uv -plane under the
(a) The magnitude of the Jacobian of the change is
(b) The integral over R xy is
RR
R xy f ( x, y ) dA xy
=
RR
R uv
∂ ( u,v )
∂ ( x,y ) f ( x (
= u, v )
∂u
∂x
∂v
∂y
− ∂u
∂y
∂v
∂x
, y ( u, v ))
.
∂ ( u,v )
∂ ( x,y )
− 1 dA uv
.
1