Matched Problem 4 Evaluate ∬ R ( y − 4 x ) d A , where R is the region in Example 4. Example 4 Evaluating a Double Integral Evaluate ∬ R ( 2 x + y ) d A , where R is the region bounded by the graphs of y = x , x + y = 2, and y = 0.
Matched Problem 4 Evaluate ∬ R ( y − 4 x ) d A , where R is the region in Example 4. Example 4 Evaluating a Double Integral Evaluate ∬ R ( 2 x + y ) d A , where R is the region bounded by the graphs of y = x , x + y = 2, and y = 0.
Solution Summary: The author evaluates the value of the iterated integral -7720.
Matched Problem 4 Evaluate
∬
R
(
y
−
4
x
)
d
A
, where R is the region in Example 4.
Example 4 Evaluating a Double Integral Evaluate
∬
R
(
2
x
+
y
)
d
A
, where R is the region bounded by the graphs of
y
=
x
, x + y = 2, and y = 0.
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.
EXAMPLE 3
Evaluate
xy dA, where D is the region bounded by the line y = x - 1 and the parabola
-3
y2 = 2x + 6.
SOLUTION
The region D is shown in the figure. Again D is both type I and type II, but the description of D
as a type I region is more complicated because the lower boundary consists of two parts. Therefore we
prefer to express D as a type II region:
-1
{ex.v) |-2 5 y s 4, - 3 sxsy+1}.
-6
-2
2
Then this equation gives
ху
xy dx dy
J-2J1/2y² - 3
Video Example )
y +1
dy
- ly•ar -(
+ 1)2
dy
+ 4y3 + 2y2 - 8y) dy
+*+ - ay",
If we had expressed D as a type I region, then we would have obtained
2x + 6
/ 2x + 6
xy dA =
xy dy dx +
xy dy dx
-V 2x + 6
but this would have involved more work than the other method.
Need Help?
Talk to a Tutor
Read It
7. Use a Jacobian coordinate switch to evaluate f (x²-y²)dA, where dA is the region bounded by
the parallelogram;
x+y=0,x+y=1,x-y=0,x-y=2
Work Problem 1 [With integrity
statement]
Let
D = {(x, y) | 2
Chapter 7 Solutions
Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.