double integrals
DESCRIPTION
intermadiate analysis double integral sectionTRANSCRIPT
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15.3 Double Integrals over General Region
Evaluating Double Integrals over General (Non-Rectangular) Regions
We will consider two types of regions
1
Type I region
A region bounded by
the lines ,x a x b= = and
the curves 1 2( ), ( )y g x y g x= = with [ ]1 2( ) ( ) ,g x g x x a b
Type I region is
1 2{( , ) : , ( ) ( )VR x y a x b g x y g x=
2 ( )y g x=
1( )y g x= X
Type II region
A region bounded by
the lines ,y c y d= = and
the curves 1 2( ), ( )x h y x h y= = with [ ]1 2( ) ( ) ,h y h y y c d
Type II region is
1 2{( , ) : , ( ) ( )}HR x y c y d h y x h y=
x = a x = b
X
Y
2 ( )x h y=
y c=
y d=
1( )x h y=
R
R
-
2
If R is type I then
2
1
( )
( )( , ) ( , )
b g x
a g xR
f x y dA f x y dydx= If R is type II then
2
1
( )
( )( , ) ( , )
d h y
c h yR
f x y dA f x y dxdy=
Step 2. Move the line left and then right. Leftmost position where the line intersects the
region R is x a= which is lower limit of the outer integral. Rightmost position where the line intersects the region R is x b= which is the upper limit of the outer integral.
Find limit of Integrats for type II region R
Step 1. Draw a horizontal line through the region R at an arbitrary fixed value y. This
line intersects the region left at the curve 1( )y h x= and right at the curve . 2 ( )y h x=Then is the lower limit and 1( )y h x= 1( )y h x= is the upper limit of the inner integral.
Step 2. Move the line down and then up. Lowest position where the line intersects the
region R is which is lower limit of the outer integral. Highest position where the
line intersects the region R is which is the upper limit of the outer integral.
y c=y d=
Step 1. Draw a vertical line through the region R at an arbitrary fixed value x. This line
intersects the region below at the curve 1( )y g x= and above at the curve . 2 ( )y g x=Then is the lower limit and 1( )y g x= 1( )y g x= is the upper limit of the inner integral.
Find limit of Integrals for type I region R
Double integrals over both regions
evaluated as iterated integrals
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Important points to note:
Outer limits always constant The idea of using a vertical line as explained in class Sometimes need to break into more integrals.
Question 5/1002: Evaluate /2 cos
sin
0 0
e drd
.
Question 9/1002: Evaluate the double integral 22
1R
y dAx +
where {( , ) : 0 1,0 }R x y x y x= .
Question 17/1002: Evaluate the double integral (2 )R
x y dA ; R is bounded by the circle with center the origin and radius 2.
Question 22/1002: Use double integral to find the volume of the solidenclosed by
the paraboloid 2 3 2z x y= + and the plane 0, 1, , 0x y y x z= = = = .
3
-
4
2
Question 28/1002: Find the volume of the region bounded by the cylinders
2 2x y r+ = and 2 2 2y z r+ = .
Question 42/1003: Sketch the region of integration and change the order of the
integration 1 /4
0 arctan
( , )f x y dydx
.
Question 48/1003: Evaluate the integral 4
3
8 2
0
x
y
e dxdy by reversing the order of integration.