Math 126 Homework hints
Note: As I’m sure you’ve all seen in section, I make mistakes too! I can’t guarantee that everything
here will be right 100% of the time. If you think there’s an error in something, please e-mail me
and I’ll try to get a correction up. Same goes for any clarification or further questions; if I’ve made
something even more confusing, please e-mail or come to office hours to clear it up!
15.2
Problem 5: Calculate the double integral.
Z Z
2(1 + x2 )
dA, R = {(x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ 1}
2
R 1+y
This integral breaks apart very easily. Since we just have a product of a term in x and a term in
y, we can factor one of them out from whichever variable we integrate first:
Z 1
Z 1
Z 2Z 1
dy
2(1 + x2 )
2
2(1 + x )
dy dx =
dx
2
1 + y2
0 1+y
0
0
0
Z 1
Z 1
dy
2
=
2(1 + x ) dx
2
0
0 1+y
Both of these integrals are not bad at all to compute, so I’ll stop here (hint: the second antiderivative
is arctan y).
Problem 7: Calculate the double integral.
Z Z
2x
dA, R = [0, 2] × [0, 1]
R 1 + xy
One thing that happens as you do double integrals more and more is that you develop an intuition
for which order of variables will be easier to integrate. As we’ve seen on the worksheets and
homework, sometimes one order may be much more difficult, if not impossible. In this case,
integrating with respect to x first turns this integral into a huge mess (I’d recommend trying it out
quickly on your own; try the substitution u = 1 + xy and see how far you can take it). So, let’s try
y first. The inner integral is then:
Z 1
2x
dy
1
+
xy
0
Make the substitution u = 1 + xy. Then as a function of y, we have u(0) = 1 and u(1) = 1 + x.
Furthermore, we get du = xdy. See why this is so nice now?
Z 1
Z x+1
2x
2
dy =
du
1
+
xy
u
0
1
x+1
= 2 ln u
1
= 2(ln(x + 1) − ln(1))
= 2 ln(x + 1)
1
So, our double integral now reduces to:
Z
2
2
ln(x + 1) dx
0
Now this is just a Math 125 problem. Use integration by parts to get the final answer.
15.3
Problem 19: Let T be the triangle enclosed by the lines y = 0, y = 8x, and x = 2. Estimate the
value of
Z Z
6 sin4 (x + y) dA
T
using the property
Z Z
(min f (x, y))(area of T ) ≤
f (x, y) dA ≤ (max f (x, y))(area of T )
T
The formula gives us explicit bounds for the integral. We only need to find the information in the
equation above and fill in the blanks to get the right answer. First, what’s that area of T ? We only
need to find the length of its base and its height. The base is given to us; it’s just 2. Its height will
be the y-coordinate of the intersection of y = 8x with y = 2, so that’s 16. The area of the triangle
is then (4)(16)/2 = 32. All that’s left is to find the maximum and minimum of 6 sin4 (x + y). Note
that sin4 (x, y) is always nonnegative. It has a minimum value of 0 at (0, 0). This tells us that the
lower bound is 0. What about the maximum?
2