xy dA where D is the domain bounded by the curves y = â x, y = 6 â x, and y = 0. Comments: If we write dA = dydx, we
Multivariate Calculus . . . . . . . . . . . . . . . . Double Integral Problems . . . . . . . . . . . . . . . . Summer 2012 Background: Notes on Double Integration, Review of Polar Coordinates, Double Integrals in Polar Coordinates Video: Introduction to Double Integrals, Iterated Integrals, Double Integrals in Polar Coordinates 1. Sketch the domain of integration of the following double integrals: Z 1 Z y=1 Z 1 Z y=x 2 (x + y 2 ) dydx, (x + y ) dydx, (b) (a) 0
Z
y=0 1 Z x=y
(c) 0
0
Z
(x + y 2 ) dxdy,
y=x 1 Z x=1
(d) 0
x=0
(x + y 2 ) dxdy.
x=y
RR √ 2. Evaluate the double integral D xy dA where D is the domain bounded by the curves y = x, y = 6 − x, and y = 0. Comments: If we write dA = dydx, we need two double integrals to perform the evaluation. √ √ Note that the line y = 6−x intersects y = x at the point (4, 2). To see this, solve x = 6−x or x = (6 − x)2 which reduces to x2 − 13x + 36 = 0. This quadratic factorizes as (x − 4)(x − 9). Since 9 lies outside the domain, the point must be (4, 2). Hence we find that Z 4 Z y=√x Z 6 Z y=6−x ZZ xy dA = xy dydx + xy dydx. 0
D
4
y=0
y=0
If we write dA = dxdy, then there is only one double integral ZZ Z 2 Z x=6−y xy dA = xy dxdy. D
0
x=y 2
3. RR Find the values where D is the domain enclosed by the unit circle: (a) RR of the following RR 5dA, (b) x dA, (c) (3x − 4y). D D D RR RR Comments: (a) Since dA = Area(D), we have D D 5dA = 5π. RR RR (b,c) By symmetry, D x dA = D y dA = 0. So the answer for both parts is 0. RR 4. Find D ydA where D is the domain in the xy-plane bounded between the line x + y = 5 and the circle x2 + y 2 = 25. Comments: The line and circle have two intersection points at (5, 0) and (0, 5). So we need to compute ZZ Z 5 Z y=√25−x2 y dA = y dy dx D
0
y=5−x
√ 2 1 2 y= 25−x y dx 2 y=5−x
5
Z = 0
=
1 2
Z
5
0
which is an integral of a polynomial in x. 1
(25 − x2 ) − (5 − x)2 dx
5. Find the volume of the solid in the first octant bounded above by the paraboloid z = x2 + 3y 2 and below by the plane z = 0, and laterally by y = x2 and y = x. Comments:The base D of the solid in the xy-plane is bounded between the curves y = x2 and y = x. We find that Z 1 Z y=x ZZ 2 2 (x2 + 3y 2 ) dydx (x + 3y ) dA = y=x2
0
D
1
Z = 0
y=x (x2 y + y 3 ) y=x2 dx
1
Z =
2x3 − (x4 + x6 ) dx
0
which is easy to evaluate. 6. Find the volume of the solid that is common to the cylinders x2 + y 2 = 25 and x2 + z 2 = 25 that lies above the xy-plane Comments: The base D of the solid is just the domain bounded by the circle x2 + y 2 = 25 so the volume is given by 5
ZZ p Z 25 − x2 dA =
−5 5
D
Z =
−5
Z
Z
√ y= 25−x2
√ y=− 25−x2
p 25 − x2 dy dx
y=√25−x2 p 25 − x2 y dx √ 2 y=− 25−x
5
(25 − x2 ) dx
= 2 −5
which is easy to evaluate. 7. Set up only the double integral for the volume of the solid bounded above by the paraboloid z = x2 + y 2 , below by the√xy-plane, and laterally by the cylinder x2 + (y − 1)2 = 1. Z 1 Z y=1+ 1−x2 Comments: (x2 + y 2 ) dy dx. √ −1
y=1− 1−x2
8. Find the volume of the solid in the first octant bounded above by the graph of z = 9 − x2 , below by z = 0, and laterally by y 2 = 3x. Comments: The domain in the xy-plane is bounded between the parabola y 2 = 3x and the line obtained from the intersection of z = 9 − x2 with z = 0 so x = 3. Then the volume is given by ZZ Z 3 Z x=3 (9 − x2 ) dA = (9 − x2 ) dxdy D
x=y 2 /3
0 3
Z = 0
x=3 (9x − x3 /3) x=y2 /3 dy
3
Z
[(27 − 9) − (3y 2 − y 6 /34 )] dy
= 0
2
which is an easy integral to evaluate in y. In reverse order, we get the equivalent double integral ZZ
Z
(9 − x2 ) dA =
3
0
D
Z
Z
√ y= 3x
(9 − x2 ) dy dx
y=0 3
= 0
y=√3x (9 − x2 )y y=0 dx
√ (9 − x2 )(− 3x) dx 0 √ Z 3 √ (9 x − x5/2 ) dx = − 3 Z
3
=
0
which again is an straightforward integral to evaluate. √ 4 Z y= x
Z 9. Reverse the order of integration in the following double integrals: (a) Z 4 Z x=8 Z 1 Z x=ey (b) f (x, y) dxdy, (c) f (x, y) dxdy. 0
x=2y
0
Comments: Z 2 Z x=4 (a) f (x, y) dxdy, 0
y=0
x=1
Z
8 Z y=x/2
(b)
x=y 2
Z f (x, y) dydx,
0
Z πZ
f (x, y) dydx, 0
y=0
π
e Z y=1
(c)
f (x, y) dydx. 1
y=ln x
sin x dx dy. (a) Describe the domain in the xy-plane given x 0 y by the limits of integration. This domain is given in terms of “horizontal cross-sections.” (b) Write down the equivalent double integral by reversing the order of integration by giving the domain by means of “vertical cross-sections.” Z (c) Evaluate the integralZ in part (b). π Z y=x π sin x Comments: The integral in reversed order is dy dx = sin x dx = 2. x 0 y=0 0
10. Consider the double integral
3
