NAME: PRINT SOLUTIONS
Calculus IV [2443002] Midterm II
Q1]...[10 points] Consider the double integral
1 0 22y 1y
f (x, y ) dx dy
Sketch the region of integration. Soln. The limits x = 2 2y and x = 1 y tell us that the region is bounded on the right by the line x + 2y = 2 and on the left by the parabola (left half) y = 1 x2 . The limits y = 0 and y = 1 tell us the upper and lower bounds for this region. We see that the parabola and line already intersect at y = 1, so the region is drawn as shown.
. .. .. .
y
y = 1 x2 .................
. . .. .. .. .. . . . . . . . . . . . . .
.. .... ........... ....... .......... ....... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .... .
x + 2y = 2
.. . .. ...
x
Reverse the order of integration. Soln. Note that reversing the order of integration means building up the region using vertical strips. There are two dierent tops on this region; the parabola top on the left side of the y -axis, and the straight line top on the right side. Thus, we have to divide the region into two pieces along the y -axis. So our answer will be a sum of two iterated integrals as shown.
0 1 0 1x2 2
f (x, y ) dy dx +
0 0
(2x)/2
f (x, y ) dy dx
Q2]...[10 points] Consider the following polar coordinates double integral
/4 /4 0 sec
r3 dr d
Sketch the region of integration. Soln. Note that the lines = /4 and = /4 correspond to the cartesian lines y = x and y = x 1 respectively. Also the curve r = sec is just r = cos which rewrites as r cos = 1 or x = 1. [Gotta hate those trig functions!] Thus we get the following region.
. .. ..
y y = x............... . .
.
. .. .. .. ... ... ... ... ... ... . . ... ... ... ... .. .. ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... ... .. .
x=1
.. . .. ...
x
y = x
Rewrite the integral as an iterated Cartesian coordinates integral (with appropriate limits). You do NOT have to compute this integral. Soln. Remember that dA = rdrd and that the remaining r2 can be written as r2 = x2 + y 2 . Thus we get the following iterated integral.
1 0 x x
x2 + y 2 dy dx
Q3]...[20 points] Use double integrals in polar coordinates to compute the surface area of the portion of the sphere x2 + y 2 + z 2 = a2 which is above the xy -plane and which lies inside the cone z 2 = x2 + y 2 . Your answer will involve a. Soln. We saw in class that surface area of a portion (over the region R) of the graph of z = f (x, y ) is given by 2 2 fx + fy + 1 dA
R
In this case we have (by implicit dierentiation of the sphere equation) fx = x/z and (likewise) fy = y/z . We did a sphere example in class! Check out the implicit dierentiation details there! Thus,
2 2 fx + fy + 1 =
x2 + y 2 + z 2 a2 =2 z2 z
which becomes
a2 a2 r2
in polar coordinates. Now the cone equation is z 2 = r2 and this intersects the sphere when r2 + r2 = a2 or r = a/ 2. Thus, the region that we are integrating over in the plane is given by 0 2 and 0 r a/ 2. Filling all this into the area surface integral (and remembering that dA = rdr d) gives
2 0 0 a/ 2
ar dr d 2 r2 a
We use the substitution u = a2 r2 (so that rdr = du/2) to evaluate the r integral. Heres what we end up with. 2 2 r 2 ]a/ 2 = (2 )(a2 / 2 + a2 ) = 2a2 (1 1/ 2) []0 [a a 0
Q4]...[20 points] Use the method of Lagrange multipliers to nd the maximum and minimum values of the function f (x, y, z ) = xy + z 2 on the sphere x2 + y 2 + z 2 = 4. Soln. f = y , x, 2z and g = 2x, 2y, 2z so the Lagrange multiplier equations are y x 2z 4 = = = = 2x 2y 2z x2 + y 2 + z 2
So thats it for the calculus. Now we just have to keep our head with all this algebra. First of all, note that the rst two equations tell us that x = 0 precisely when y = 0 (since x and y are multiples of each other). So lets break this analysis into two cases. Case I: [x = 0 and y = 0] In this case the fourth equation becomes z 2 = 4 and so we get z = 2. Thus, we get two points: (0, 0, 2) and (0, 0, 2). Case II: [x = 0 and y = 0] In this case the rst two equations give x/y = 2 = y/x. But this means that 2 must be equal to its own reciprocal (since x/y and y/x are reciprocals) and so must be 1. Thus the 2 third equation becomes z = z which that = 0. Now equation 4 2 implies z becomes 2x = 4 or x = 2. We get four points: ( 2, 2, 0), ( 2, 2, 0), ( 2, 2, 0), and ( 2, 2, 0). Finally, we evaluate f on these 6 points and see that the maximum f -value is 4 (occurs at (0, 0, 2) and (0, 0, 2)), and that the minimum f -value is -2 (occurs at ( 2, 2, 0) and ( 2, 2, 0)). Bonus Question. Let u and v be dierentiable functions of one variable with derivatives denoted by u and v respectively. Let R be the triangular region with vertices at the points (a, a), (b, b) and (b, a). By evaluating the double integral u (x) v (y ) dA
R
in two dierent ways (as iterated integrals), give a new derivation of the integration by parts formula
b a
u dv = uv |b a
b
v du .
a
Soln.
. . .. ..
y
.. ... ... ... ... ... ... ... ... ... .. .. ... ... ... ... ... ... .. ... ... ... ... ... ... .. ... ... .. .. ... ... ... ... ... ... ... ... .. ...
b
(b, b)
Here is a diagram of the triangular region with its sides labelled.
a
y=x
x=b
(a, a) a
y=a
(b, a)
... . .. ..
b
x
On the one hand, we can integrate with respect to y rst to get
b x
u (x) v (y ) dA =
R a b a
u (x)v (y ) dy dx [u (x)v (y )]x dx a u (x)v (x) u (x)v (a) dx v (x)u (x) dx [u(x)v (a)]b a v (x)u (x) dx u(b)v (a) + u(a)v (a)
=
a b
=
a b
=
a b
=
a
On the other hand, we can integrate with respect to x rst to get
b b
u (x) v (y ) dA =
R a b y
u (x)v (y ) dx dy [u(x)v (y )]b dy y u(b)v (y ) u(y )v (y ) dy
b
=
a b
=
a
= [u(b)v (y )]b a
u(y )v (y ) dy
a b
= u(b)v (b) u(b)v (a) Finally, setting these two expressions equal to each other gives
b a
u(y )v (y ) dy
a
b
v (x)u (x) dx u(b)v (a) + u(a)v (a) = u(b)v (b) u(b)v (a)
u(y )v (y ) dy
a
which simplies down to (removing explicit reference to the dummy variables of integration x and y )
b a
v du = uv |b a
b
u dv
a
and were nished.