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TEST 1
- Rewrite the integral
− 1
x^2
∫ (^1) −y 0
f (x, y, z)d zd yd x as an iterated integral in the order d xd yd z and d yd zd x.
(A) Solid E. (B) Projection onto x z-plane. (C) Projection onto y z-plane.
∫ (^1) − 1
x^2
∫ (^1) −y 0
f (x, y, z)d zd yd x =
0
∫ (^1) −z 0
∫ py −py
f (x, y, z)d xd yd z =
− 1
∫ (^1) −x 2 0
∫ (^1) −z x^2
f (x, y, z)d yd zd x.
- Evaluate the integral
E
y zdV , where E lies above the plane z = 0, below the plane z = y, and inside the
cylinder x^2 + y^2 = 4.
E
y zdV =
x^2 +y^2 ≤4,y≥ 0
∫ (^) y 0
y zd zd A =
∫ (^) π 0
0
∫ (^) r sin θ 0
r sin θ z r d zd r d θ =
∫ (^) π 0
0
r 2 sin θ z^2 2
r sin θ
0
d r d θ
∫ (^) π 0
0
r 4 sin^3 θ d r d θ =
∫ (^) π 0
sin^3 θ r 5 5
2
0
d θ =
∫ (^) π 0
sin θ
Ä
1 − cos^2 θ
ä d θ
Ç
− cos θ + cos^3 θ 3
å (^) π
0
- Find the area of the region in the first quadrant bounded by the curves y = px, y = 8 px, and the hyperbolas xy = 1, xy = 4.
Set u = xy and v = y / p x. Then 1 ≤ u ≤ 4, 1 ≤ v ≤ 8, and x = (u / v)^2 /^3 , y =
uv^2
. Therefore, the Jacobian is equal to (^)
u−^1 /^3 v−^2 /^3 −
u^2 /^3 v−^5 /^3 1 3 u−^2 /^3 v^2 /^3
u^1 /^3 v−^1 /^3
v
v
3 v
Thus, the area of the region is equal to ∫ (^4) 1
1
3 v d vd u^ =
1
3 ln^ v
8 1
d u =
1
2 ln 2d u = 6 ln 2.
- Evaluate the integral
0
0
emax{x,y}^ d xd y, where max{x, y} means the larger of x and y.
y is bigger than x above the line y = x and x is bigger than y below this line. Hence,
emax{x,y}^ =
ey^ , in { 0 ≤ x ≤ 1, x ≤ y ≤ 1 }, e x^ , in { 0 ≤ x ≤ 1, 0 ≤ y ≤ x}.
Therefore, ∫ (^1) 0
0
emax{x,y}^ d xd y =
0
x
ey^ d yd x +
0
∫ (^) x 0
e x^ d yd x =
0
(e − e x^ )d x +
0
x e x^ d x
= (e x − e x^ )|^10 + x e x^ |^10 −
0
e x^ d x = (e − e) − (− 1 ) + e − e x^ |^10 = 1 + e − (e − 1 ) = 2.
- Use the transformation x = u^4 , y = v^4 , and z = w^4 , to evaluate the integral of
pxy z over the solid bounded by
the surface px + py + pz = 1 and the coordinate planes.
Since p x + py + p z = 1 becomes u^2 + v^2 + w^2 = 1, the initial solid is carried over into a solid E which is the part of the sphere u^2 + v^2 + w^2 ≤ 1 lying in the first octant. The Jacobian of this transformation is equal to ^4 u
0 4 v^3 0 0 4 w^3
= 64 (uvw) (^3).
As (xy z)−^1 /^2 = (uvw)−^2 , the initial integral is equal to
64
E
uvwd wd vd u = 64
∫ (^) π/ 2 0
∫ (^) π/ 2 0
0
ρ sin φ cos θρ sin φ sin θρ cos φρ^2 sin φ d ρ d φ d θ
= 64
∫ (^) π/ 2 0
cos θ sin θ d θ
∫ (^) π/ 2 0
sin^3 φ cos φ d φ
0
ρ^5 d ρ
sin^2 θ 2
π/ 2
0
sin^4 φ 4
π/ 2
0
ρ^6 6
1
0
