Module MA1132 (Frolov), Advanced Calculus Homework Sheet 7 Each set of homework questions is worth 100 marks Due: at the beginning of the tutorial session Thursday/Friday, 17/18 March 2016 Name: You may use Mathematica to sketch the integration regions and solids, and to check the results of integration. 1. Evaluate the double integral I = ZZ R 1 x + y dxdy , where R is the region enclosed by the lines x = 2, y = x, and the hyperbola xy = 1. Solution: The region R is shown below 0.0 0.5 1.0 1.5 2.0 0.0 0.5 1.0 1.5 2.0 We consider R as a region of type I. Then I is given by I = Z 2 1 Z x 1/x 1 x + y dy dx = Z 2 1 ln(x + y)| y=x y=1/x dx = Z 2 1 ln( 2x 2 1+ x 2 ) dx = ln( 2x 2 1+ x 2 )x 2 1 - Z 2 1 xd ln( 2x 2 1+ x 2 ) = 2 ln 8 5 - Z 2 1 2 1+ x 2 dx = 2 ln 8 5 + π 2 - 2 arctan 2 ≈ 0.296506 . (1) 2. Sketch the integration region R and reverse the order of integration (a) Z 7/2 -1/2 Z 2+ √ 7+12y-4y 2 2- √ 7+12y-4y 2 f (x, y)dxdy (2) 1
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Due: at the beginning of the tutorial session Thursday/Friday, 17/18 March 2016
Name:
You may use Mathematica to sketch the integration regions and solids, and to check theresults of integration.
1. Evaluate the double integral
I =
∫∫R
1
x+ ydxdy ,
where R is the region enclosed by the lines x = 2, y = x, and the hyperbola xy = 1.
Solution: The region R is shown below
0.0 0.5 1.0 1.5 2.0
0.0
0.5
1.0
1.5
2.0
We consider R as a region of type I. Then I is given by
I =
∫ 2
1
[∫ x
1/x
1
x+ ydy
]dx =
∫ 2
1
ln(x+ y)|y=xy=1/x dx =
∫ 2
1
ln(2x2
1 + x2) dx
= ln(2x2
1 + x2)x
∣∣∣∣21
−∫ 2
1
x d ln(2x2
1 + x2) = 2 ln
8
5−∫ 2
1
2
1 + x2dx
= 2 ln8
5+π
2− 2 arctan 2 ≈ 0.296506 .
(1)
2. Sketch the integration region R and reverse the order of integration
(a) ∫ 7/2
−1/2
∫ 2+√
7+12y−4y2
2−√
7+12y−4y2f(x, y)dxdy (2)
1
Solution: The region R is shown below
-2 0 2 4 6
0
1
2
3
It is found by noting that
2−√
7 + 12y − 4y2 ≤ x ≤ 2 +√
7 + 12y − 4y2 =⇒ (x− 2)2 ≤ 7 + 12y − 4y2
=⇒ (x− 2)2 + 4(y − 3
2)2 ≤ 16 =⇒ (x− 2)2
16+
(y − 32)2
4≤ 1 .
(3)Thus it is a closed region bounded by an ellipse centred at (2, 3/2) with semi-axis 4and 2. Reversing the order of integration one gets
∫ 7/2
−1/2
∫ 2+√
7+12y−4y2
2−√
7+12y−4y2f(x, y)dxdy =
∫ 6
−2
∫ 32+
√3+x−x2
4
32−√
3+x−x2
4
f(x, y)dydx . (4)
(b) ∫ 1
0
∫ √3−x2x2/2
f(x, y)dydx (5)
Solution: The region R is shown below
0.0 0.2 0.4 0.6 0.8 1.0
0.0
0.5
1.0
1.5
2
Reversing the order of integration one gets the sum of three repeated integrals∫ 1
0
∫ √3−x2x2/2
f(x, y)dydx
=
∫ 1/2
0
∫ √2y0
f(x, y)dxdy +
∫ √21/2
∫ 1
0
f(x, y)dxdy +
∫ √3√2
∫ √3−y2
0
f(x, y)dxdy .
(6)
3. Sketch the integration region R and write the expression as one repeated integral byreversing the order of integration∫ 6
2
∫ 3
9/(x+1)
f(x, y)dydx+
∫ 8
6
∫ 9−x
9/(x+1)
f(x, y)dydx (7)
Solution: The region R is shown below
2 3 4 5 6 7 81.0
1.5
2.0
2.5
3.0
Reversing the order of integration one gets∫ 6
2
∫ 3
9/(x+1)
f(x, y)dydx+
∫ 8
6
∫ 9−x
9/(x+1)
f(x, y)dydx =
∫ 3
1
∫ 9−y
−1+9/y
f(x, y)dxdy . (8)
4. Prove the Dirichlet formula∫ b
a
∫ x
a
f(x, y)dydx =
∫ b
a
∫ b
y
f(x, y)dxdy , (9)
and use it to prove that∫ x
a
∫ t1
a
(t1 − t)n−1f(t)dtdt1 =1
n
∫ x
a
(x− t)nf(t)dt . (10)
Solution: The Dirichlet formula follows from the fact that the repeated integrals are equalto the double integral over the triangle enclosed by the lines y = a, x = b, y = x.
Thus, we get∫ x
a
∫ t1
a
(t1 − t)n−1f(t)dtdt1 =
∫ x
a
∫ x
t
(t1 − t)n−1f(t)dt1dt
=
∫ x
a
1
n(t1 − t)n|t1=xt1=t f(t)dt =
1
n
∫ x
a
(x− t)nf(t)dt .
(11)
3
5. Find the volume V of the solid bounded by
(a) the planes x = 0, y = 0, z = 0, 2x+ y = 1, and the surface z = x+ y2 + 1.
Solution: The solid, and its projection R onto the xy-plane are shown below
0.0 0.1 0.2 0.3 0.4 0.5
0.0
0.2
0.4
0.6
0.8
1.0
Thus, the volume is
V =
∫∫R
(x+ y2 + 1)dxdy =
∫ 1
0
∫ 1−y2
0
(x+ y2 + 1)dxdy
=
∫ 1
0
(1
8(1− y)2 +
1
2(y2 + 1)(1− y))dy =
1
8
∫ 1
0
(−4y3 + 5y2 − 6y + 5)dy
=1
8(−1 +
5
3− 3 + 5) =
1
3.
(12)
(b) the planes x = 1, z = 0, the parabolic cylinder x − y2 = 0, and the paraboloidz = x2 + y2.
Solution: The solid, and its projection R onto the xy-plane are shown below
0.0 0.2 0.4 0.6 0.8 1.0
-1.0
-0.5
0.0
0.5
1.0
Thus, the volume is
V =
∫∫R
(x2 + y2)dxdy =
∫ 1
−1
∫ 1
y2(x2 + y2)dxdy =
∫ 1
−1(y2(1− y2) +
1
3(1− y6))dy
= 2
∫ 1
0
(y2 − y4 +1
3− 1
3y6)dy = 2(
1
3− 1
5+
1
3− 1
21) =
88
105.
(13)
4
(c) the planes x = 0, y = 0, z = 0, the cylinders az = x2, a > 0, x2 + y2 = b2, andlocated in the first octant x ≥ 0, y ≥ 0, z ≥ 0.
Solution: The solid, and its projection R onto the xy-plane are shown below (a = 2, b = 1)
Thus, the volume is
V =
∫∫R
1
ax2dxdy =
1
a
∫ b
0
∫ √b2−x20
x2dydx =1
a
∫ b
0
x2√b2 − x2dx
=b4
a
∫ 1
0
x2√
1− x2dx =b4
a
∫ π/2
0
sin2 t cos2 tdt =b4
a
∫ π/2
0
1
4sin2 2tdt
=b4
4a
∫ π/2
0
1
2(1− cos 4t)dt =
πb4
16a.
(14)
(d) the planes z = zmin, z = λx+ µy + h (λ > 0, µ > 0, h > 0), and the elliptic cylinderx2
a2+ y2
b2= 1, where zmin is the z-coordinate of the lowest point on the intersection of
z = λx+ µy + h (λ > 0, µ > 0, h > 0) and x2
a2+ y2
b2= 1.
Solution: The intersection, the solid, and its projection R onto the plane z = zmin are shownbelow (a = 1, b = 2, λ = 1, µ = 2, h = 5)
-1.0 -0.5 0.0 0.5 1.0
-2
-1
0
1
2
5
To find zmin we use the Lagrange multiplier method, and get the equations
λ = 2Λx
a2, µ = 2Λ
y
b2,
x2
a2+y2
b2= 1 . (15)
Multiplying the first equation by x, the second equation by y and summing up theresulting equations, one gets
λx+ µy = 2Λ . (16)
Thus, one derives the equations for x and y
λ = (λx+ µy)x
a2, µ = (λx+ µy)
y
b2, (17)
or by using the constraint x2
a2+ y2
b2= 1
y2
b2=
µ
λa2xy ,
x2
a2=
λ
µb2xy . (18)
Summing up these equations one finds
xy =λµa2b2
λ2a2 + µ2b2, x2 =
λ2a4
λ2a2 + µ2b2, y2 =
µ2b4
λ2a2 + µ2b2. (19)
Thus there are two solutions
xmin = − λa2√λ2a2 + µ2b2
, ymin = − µb2√λ2a2 + µ2b2
,
xmax = +λa2√
λ2a2 + µ2b2, ymax = +
µb2√λ2a2 + µ2b2
,
(20)
corresponding to the lowest and highest points on the intersection, respectively. Thez-coordinates of these points are
zmin = h−√λ2a2 + µ2b2 , zmax = h+
√λ2a2 + µ2b2 . (21)
To find the volume we first shift the z-coordinate z → z + zmin so that the planez = zmin becomes the plane z = 0, and the plane z = λx+ µy + h becomes