Math 2210–001, Midterm 3, Study Guide
Summer 2003
(1) If w = x2 y + z 2 , x = ρ cos θ sin φ, y = ρ sin θ sin φ, and z = ρ cos φ, then find
∂w .
∂θ ρ=2,θ=π,φ=π/2
Solution: First off,
∂x
= −ρ sin θ cos φ
∂θ
∂y
= ρ cos θ sin φ
∂θ
∂z
= 0.
∂θ
∂w
= 2xy = 2ρ cos θ sin φ
∂x
∂w
= x2 = ρ2 cos2 θ sin2 φ
∂y
∂w
= 2z = 2ρ cos φ
∂z
Therefore,
∂w ∂x ∂w ∂y ∂w ∂z
∂w
=
·
+
·
+
·
∂θ
∂x ∂θ
∂y ∂θ
∂z ∂θ
= −2ρ3 cos θ sin θ sin φ cos φ + ρ3 cos3 θ sin3 φ
= ρ3 cos θ sin φ −2 sin θ cos φ + cos2 θ sin2 φ .
When ρ = 2, θ = π, and φ = π/2, the above is equal to −8.
(2) Find the minimum of f (x, y) = x2 + y 2 subject to x + y = 0.
Solution: x + y = 0 means that y = −x, so the function, along this line, is f (x, −x) = 2x2 . Its minimum is zero at
x = 0. However, you can also use Lagrange multipliers as follows: Let g(x, y) = x+ y, and solve ∇f (x, y) = λ∇g(x, y)
subject to g(x, y) = 0. Now ∇f (x, y) = 2xi + 2yj, and ∇g(x, y) = i + j. So ∇f = λ∇g if and only if 2x = λ and
2y = λ; i.e.,
λ
(Eq-1)
x=y= .
2
But now the condition g(x, y) = 0 tells us that x = −y, so that, by (Eq-1), λ = −λ. The only solution to the latter
equation is λ = 0. Hence, at the minimum, x = y = 0, and f (x, y) = 0.
R1R1
2
2
(3) What is the numerical value of 0 0 xyex +y dx dy?
Solution: Note that
Z
1
xyex
2
+y 2
dx = yey
0
2
Z
1
2
xex dx =
0
1 y2
ye
2
Z
1
er dr
(r = x2 )
0
1 y2
ye (1 − e).
2
This is the inside integral. Plug it into the outside integral to get
Z 1
Z 1Z 1
2
2
2
1
xyex +y dx dy = (1 − e)
yey dy
2
0
0
0
1
= (1 − e)2 .
4
=
1
(s = y 2 )
2
(4) (4 points each) Consider the surface that is described by z = x3 − 3yz 2 .
(a) Compute the equation of the tangent plane at any point (x0 , y0 , z0 ).
Solution: This surface is best described, implicitly, by F (x, y, z) = 0, where
F (x, y, z) = x3 − 3yz 2 − z.
First, note that
∇F (x, y, z) = 3x2 i − 3z 2 j − (6yz + 1)k.
But the tangent plane at (x0 , y0 , z0 ) is described by the set of points (x, y, z) such that ∇F (x0 , y0 , z0 ) · hx −
x0 , y − y0 , z − z0 i = 0; i.e.,
3x20 (x − x0 ) − 3z02 (y − y0 ) − (6y0 z0 + 1)(z − z0 ) = 0.
(Eq-2)
(b) Find all points on this surface where the tangent plane is horizontal.
Solution: We can rewrite (Eq-2) as Ax + By + Cz = D, where A = 3x20 , B = −3z02, C = −(6y0 z0 + 1), and
D = 3x30 − 3z02 y0 + (6y0 z0 + 1)z0 . So this plane is horizontal if and only if C = 0; i.e., if 6y0 z0 + 1 = 0. That is,
the tangent plane at (x0 , y0 , z0 ) is horizontal if and only if y0 z0 = − 61 .
(5) What is the volume of the solid that is: (a) in the first octant; (b) and bounded by the circular cylinders x2 + z 2 = 4
and y 2 + z 2 = 4?
Solution: The numerical answer to this problem is 4π. To do this, we first plot the surfaces to find that:
(a) They meet if and only if x2 + z 2 = 4 and y 2 + z 2 = 4, so that by setting the z 2 ’s equal you get x = y (in the
first octant).
(b) When x < y, the surface x2 + z 2 = 4 is above the other.
(c) When x > y, the surface x2 + y 2 = 4 is below the other.
(d) All of the action, on the xy-plane, takes place on the square 0 ≤ x ≤ 2 and 0 ≤ y ≤ 2.
Assuming that the object has constant density 1 everywhere, the volume, for the portion x < y, is the volume under
z 2 = 4 − x2 minus the volume under z 2 = 4 − y 2 ; i.e.,
Z 2Z yp
Z 2Z 2p
4 − x2 dy dx −
4 − y 2 dx dy
0
(Eq-3)
Z
x
2
=
0
0
Z
p
(2 − x) 4 − x2 dx −
0
0
2
y
p
4 − y 2 dy.
The full volume, and hence the answer to the question, is twice this by symmetry. Now leaving the integrals intact
is perfectly acceptable for the exam, but let us compute them here. The second integral is easier:
Z
Z 2 p
1 4√
2
y 4 − y dy =
u du
(u = 4 − y 2 )
2
0
0
(Eq-4)
8
= .
3
As to the first integral in (Eq-3),
Z 2p
Z 2 p
Z 2
p
2
2
(2 − x) 4 − x dx = 2
4 − x dx −
x 4 − x2 dx
0
0
Z
=8
0
1
0
p
8
1 − u2 du −
3
(u = x/2).
3
√
√
The second term follows from (Eq-4). Now the antiderivative of 1 − u2 is 12 u 1 − u2 + 12 sin−1 u + C. So
R1√
1 − u2 du = 12 sin−1 (1) = π4 . Thus, the first term in (Eq-3) is 2π − 83 . So the volume over the region x < y is
0
2π − 83 + 83 = 2π. The full volume is twice this; i.e., 4π.