MHB Using Reimann sum to estimate the value of a double integral

Click For Summary
To estimate the value of the double integral ∫∫R(y² - 2x²) dA over the region R = [−3, 1] × [−2, 0] using a Riemann sum with m = 4 and n = 2, the upper left corners of the squares should be used as sample points. The initial steps included finding the indefinite integral, which is y³/3 - 2x³/3. However, the focus should remain on applying the Riemann sum method rather than integrating directly. The importance of using Riemann sums for estimation was emphasized, indicating a misunderstanding of the task. The discussion highlights the need to clarify the approach to solving the problem correctly.
carl123
Messages
55
Reaction score
0
If R = [−3, 1] × [−2, 0], Use a Riemann sum with m = 4, n = 2 to estimate the value of ∫∫R(y2 − 2x2) dA. Take the sample points to be the upper left corners of the squares.

So far,

I found the indefinite integral of the function to be y3/3 - 2x3/3

Not sure where to go from here
 
Physics news on Phys.org
carl123 said:
If R = [−3, 1] × [−2, 0], Use a Riemann sum with m = 4, n = 2 to estimate the value of ∫∫R(y2 − 2x2) dA. Take the sample points to be the upper left corners of the squares.

So far,

I found the indefinite integral of the function to be y3/3 - 2x3/3

Not sure where to go from here

Why are you integrating at all? You were told to use Riemann Sums to ESTIMATE the integral!
 
Thread 'Problem with calculating projections of curl using rotation of contour'
Hello! I tried to calculate projections of curl using rotation of coordinate system but I encountered with following problem. Given: ##rot_xA=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}=0## ##rot_yA=\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}=1## ##rot_zA=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y}=0## I rotated ##yz##-plane of this coordinate system by an angle ##45## degrees about ##x##-axis and used rotation matrix to...

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K