MATH 16B - W ORKSHEET 3 - S OLUTIONS
1 Compute all first order partial derivatives for the following functions
i f (x, y) =
ln(xy)
sin(x+y)
A Apply the quotient rule to obtain
@f
@x
=
1
sin(x+y) x
ln(xy) cos(x+y)
sin(x+y)2
and
@f
@y
=
sin(x+y) y1 ln(xy) cos(x+y)
.
sin(x+y)2
⌅
ii f (x, y) = cos(ex y + 1)
A Apply the chain rule to obtain
iii f (x, y) = xz(xy
@f
@x
sin(ex y + 1) · (ex y) and
=
@f
@y
sin(ex y + 1) · (ex ). ⌅
=
exy )
A Since xz is a constant with respect to y, we have
x(xy exy ).
@f
@y
= xz ·
@(xy exy )
@y
= xz(x
xexy ). Similarly,
exy ) + xz(y
yexy ) = 2xyz
@f
@z
=
@(xz)
@z
· (xy
For the last partial, apply the product rule to obtain
@f
@(xz)
=
· (xy
@x
@x
exy ) + (xz) ·
@
(xy
@x
exy ) = z(xy
(z + xyz)exy . ⌅
2
i Compute
ˆ ˆ
(x2 + y 2 )dxdy where D is the rectangle bounded by the lines x = 1, x = 3, y = 0 and y = 1.
D
A Since D is just a rectangle, we can directly compute this:
ˆ
0
1
ˆ
3
(x2 + y 2 )dxdy
=
1
ˆ
1
ˆ
1
0
=
=
=
=
1
✓
✓
1 3
x + y2 x
3
◆
3
dy
1
◆
1
2
9 + 3y
y dy
3
0
◆
ˆ 1✓
26
2y 2 +
dy
3
0
2 3 26
y + y
3
3
28
.⌅
3
2
1
0
exy ) =
ii Compute the following double integral
x=
ˆ ˆ
(x + y)2 dxdy, where R is the region enclosed by y = x, y = 4 and
R
1.
A
There are two ways to parameterize the region R: R = { 1  x  4, x  y  4} or R = { 1  y  4, 1  x  y}.
ˆ 4ˆ 4
In the first case (diagram on the left) we have sliced vertically, so our integral is of the form
(x + y)2 dydx. In
1 x
ˆ 4ˆ y
the second case (diagram on the right) we have sliced horizontally, so our integral is of the form
(x + y)2 dxdy.
1
1
These two expressions, once evaluated, will give the same number. Thus, we can use our second parameterization
and compute the integral:
ˆ ˆ
2
(x + y) dxdy
=
R
=
=
=
=
=
=
ˆ
ˆ
ˆ
1
3
4
1
4
1
4
1
✓
ˆ
y
(x + y)2 dxdy
1
1
(x + y)3
3
1
8y 3
3
8 4
y
4
✓
1 8 4
·4
3 4
1
(8 · 44
12
1975
.⌅
12
2
y
(y
1
(y
4
1
(4
4
34
dy
1
1)3 dy
1)4
◆
1)
4
4
8 + 24 )
1
8
1
( 1)4 + ( 2)4
4
4
◆
iii Compute the area under the graph y =
+ x2 between x = 1 and x = 3.
1
x
A Recall that the integral of a “non-negative” valued function f (x) from a to b, by definition, is the area under the graph
of y = f (x) from a to b. Since x1 + x2 0 for all x, the area is just
ˆ
1
iv Compute
ˆ ˆ
3
✓
◆
✓
◆
1
x3
2
+ x dx = ln x +
x
3
3
= ln 3 +
27
1
3
1
= ln 3 +
xydxdy where D is the region bounded by y = x3 and y =
p
x.
D
A Here’s what the region looks like:
The two curves intersect when x3 =
p
x i.e when x = 0, 1. Thus, our integral is just,
ˆ
0
1
ˆ
p
x
xydydx =
x3
ˆ
1
ˆ
1
0
=
0
=
=
=
3
✓
y2
x
2
x2
2
x3
2·3
1
1
6 16
5
.⌅
48
p
x
dx
x3
x7
dx
2
◆
x8
2·8
1
0
26
.⌅
3