
16-16 Calculus – Differentiation and Integration
= − −
= − − +
= − +
e y e
y
e e
e e
y
( )
( )
.
1
2
2 1 2 1
2 1
2
0
2
2
2
Example 16.16 Evaluate f x y dA
R
( , )
òò
for f(x, y) = x
2
y
2
, where R is the region in the first quadrant
enclosed by x = 0, y = 0 and x
2
+ y
2
= 1.
Solution: Here,
R x y x y x= ≤ ≤ ≤ ≤ −
( , ) : , .0 1 0 1
2
f x y dA x y dy dx
x y
R
x
( , )
∫∫
=
∫
∫
=
−
−
2 2
0
1
0
1
2 3
0
1
2
3
xx
dx
x x dx
2
0
1
2 2 3 2
0
1
1
3
1
∫
= −
∫
( )
/
Substitute x = sin q, then dx = cos q dq.
=
∫
=
∫
1
3
1
3
2
0
2
2 3 2
2
0
2
4
sin (cos ) (cos )
sin (cos )
/
/
/
θ θ θ θ
θ θ θ
π
π
d
d
Figure 16.5
x
y
x
2
+ y
2
= 1
y = 0
x
= 0
R