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(MATH100)final-rev-sol.pdf
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MATH 100 (L1) Introduction to Multivariable Calculus Spring 2006-07
. Final Review Solutions (May 10, 2007)
.. 1
4.1 (Double Integrals) Evaluate x cos(xy) sin2 x dA, R = {(x, y):0 . x . , 0 . y . }.
2
R
. 1/2 . . 1/2 . 1/2 . I = x cos(xy) sin2 x dydx = sin2 x . sin(xy). dx = sin3 x dx 000 0
y=0
.1 . 1/2 .1 . cos3 x .1/2 .1 .1 . 2
= (1 . cos2 x) d(cos x) = cos x . = (0 . 0) . (1 . )= .
0 3 33
0
Remark that the double integral was expressed as an iterated integral for which the ranges of integration are all constants. In this case we may interchange the order of integration. Let us take a look if we really do this.
. 1/2 . . . 1/2 I = x cos(xy) sin2 x dydx = x cos(xy) sin2 x dx dy. 00 00
We observe that it is much easier to evaluate the .rst iterated integral (just as we did) than the second one. Students however should be reminded that if the domain of integration is only a rectangular region, then the order of integration can freely be reversed. Sometimes this will enable us to turn a di.cult integral into an easier one.
. 2 . e
x

4.2 (Double Integrals over Non-rectangular Regions) (a) Evaluate xy .1 dy dx.
11/x
y
(b) Use double integral to .nd the volume bounded by z = sin , z = 0, y = x, y = 0, x = .
2x
. 2 . e . 2 . 2 . (a) x x y .1 dy dx = x x. ln y.e dx = x x(1 + ln x) dx = .x x.2 = 3. 11/x 11
y=1/x 1
Note that the anti-derivative of xx (1+ln x) is only xx+C because d (x x)= x x(1+ln x),
dx
by logarithmic di.erentiation.
. (b) The domain of integration R is sketched as follows:
y y = x
x

.. y . . x y . ..2x y .x
Volume = sin dA = sin dy dx = cos dx
2x 2x 2x
00 0 y=0 R
. .2x 2 . 2 2
= (0 . 1) dx = x dx = ()= .
0 0 2
.. y . . y
There really is nothing wrong to write the volume as sin dA = sin dx dy, but
2x 2x
0 y R
however the latter one has an unsolvable inner integral.
4.3 (Reverse Order of Integration) Express the followings as double integrals with the order of inte-gration reversed.

2ax
. 2 . 2.x . 2a .
(a) f(x, y) dy dx, (b) f(x, y) dydx, a > 0.

2
.6 x2/4.1 02ax.x
. The domains of integration are sketched as follows:
y 2 y
x
y = . 1
y =2ax
4
8 2a
a
y =2ax . x2
x
.6 .12 x
02a y =2 . x
(a) (b)

. 0 . 2 y+1 . 8 . 2.y
(a) I1 = f(x, y) dx dy + f(x, y) dx dy.

.1 .2 y+1 0 .2 y+1

2
. a . a. a2.y. a . 2a . 2a . 2a
(b) I2 = f(x, y) dx dy + f(x, y) dx dy + f(x, y) dx dy.
2
0 y2/(2a)0 a+ a2.yay2/(2a)
4.4 (Polar Coordinates) Use polar coordinates to evaluate the double integral
sin .x2 + y2 dx dy. 2.x2+y2.42
. Using polar coordinates x = r cos and y = r sin , where . r . 2 and 0 . . 2.
y r = r =2
x
. 2 . 2 . 2 ..
2 . 2 .
I = sin r2 r dr d = (2) r sin r dr = (2) .r cos r+ cos r dr= .62 .

0
4.5 (Surface Area) Find the total