The formula obtained in part (b) of Exercise 43 is useful in integration problems where it is inconvenient or impossible to solve the transformation equations u = f x , y , υ = g x , y explicitly for x and y in terms of u and υ . In these exercises, use the relationship ∂ x , y ∂ u , υ = 1 ∂ u , υ / ∂ x , y to avoid solving for x and y in terms of u and υ . Use the transformation u = x y , υ = x y 4 to find ∬ R sin x y d A where R is the region enclosed by the curves x y = π , x y = 2 π , x y 4 = 1 , x y 4 = 2.
The formula obtained in part (b) of Exercise 43 is useful in integration problems where it is inconvenient or impossible to solve the transformation equations u = f x , y , υ = g x , y explicitly for x and y in terms of u and υ . In these exercises, use the relationship ∂ x , y ∂ u , υ = 1 ∂ u , υ / ∂ x , y to avoid solving for x and y in terms of u and υ . Use the transformation u = x y , υ = x y 4 to find ∬ R sin x y d A where R is the region enclosed by the curves x y = π , x y = 2 π , x y 4 = 1 , x y 4 = 2.
The formula obtained in part (b) of Exercise 43 is useful in integration problems where it is inconvenient or impossible to solve the transformation equations
u
=
f
x
,
y
,
υ
=
g
x
,
y
explicitly for x and y in terms of
u
and
υ
.
In these exercises, use the relationship
∂
x
,
y
∂
u
,
υ
=
1
∂
u
,
υ
/
∂
x
,
y
to avoid solving for x and y in terms of
u
and
υ
.
Use the transformation
u
=
x
y
,
υ
=
x
y
4
to find
∬
R
sin
x
y
d
A
where R is the region enclosed by the curves
x
y
=
π
,
x
y
=
2
π
,
x
y
4
=
1
,
x
y
4
=
2.
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
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