The double integral of f over rectangle R and midpoint rule for double integrals

[Petrus]

Hello MHB,
I wanted to 'challange' myself with solve a problem with midpoint and rule and the double integral f over the rectangle R.
This is a problem from midpoint.
"Use the Midpoint Rule m=n=2 to estimate the value of the integrab $$\int\int_r(x-3y^2)dA$$, where $$R= {(x,y)| 0\leq x \leq 2, 1 \leq y \leq 2}.$$
Let's start with mid point. I Always start with draw itView attachment 738
and the center of those four subrectangles is $$x_1=\frac{1}{2}$$, $$x_2= \frac{3}{2}$$, $$y_1=\frac{5}{4}$$ and $$y_2=\frac{7}{4}$$ and the area of the sub rectangle is $$A= \frac{1}{2}$$ to make it more clear we got $$\frac{1}{2}(f(\frac{1}{2},\frac{5}{4})+f(\frac{1}{2},\frac{7}{4})+f(\frac{3}{2},\frac{5}{4})+f(\frac{3}{2},\frac{7}{4})=-11.875$$
this question is aproxomite same as
$$\int_1^2\int_0^2 x-3y^2 dxdy$$ and that will give result $$-13$$
Now for the most chalange one that I can't solve
I can only solve it with $$m=2$$$$n=1$$(we got two subs in x and 1 in y. cause then we get I start as usually draw it :) View attachment 739 then we got the area of sub square $$A=1$$ so we got these point $$f(1,2)+f(2,2)$$
and we get result $$-9$$. Now afterwards I type I notice we can't use the rule right? $$x-3y^2=z$$ and for any of our point we will get $$z<0$$ and then we can't use defination. but are my two above correct?Regards,

 

[Sudharaka]

Hello MHB,
I wanted to 'challange' myself with solve a problem with midpoint and rule and the double integral f over the rectangle R.
This is a problem from midpoint.
"Use the Midpoint Rule m=n=2 to estimate the value of the integrab $$\int\int_r(x-3y^2)dA$$, where $$R= {(x,y)| 0\leq x \leq 2, 1 \leq y \leq 2}.$$
Let's start with mid point. I Always start with draw itView attachment 738
and the center of those four subrectangles is $$x_1=\frac{1}{2}$$, $$x_2= \frac{3}{2}$$, $$y_1=\frac{5}{4}$$ and $$y_2=\frac{7}{4}$$ and the area of the sub rectangle is $$A= \frac{1}{2}$$ to make it more clear we got $$\frac{1}{2}(f(\frac{1}{2},\frac{5}{4})+f(\frac{1}{2},\frac{7}{4})+f(\frac{3}{2},\frac{5}{4})+f(\frac{3}{2},\frac{7}{4})=-11.875$$
this question is aproxomite same as
$$\int_1^2\int_0^2 x-3y^2 dxdy$$ and that will give result $$-13$$
Now for the most chalange one that I can't solve
I can only solve it with $$m=2$$$$n=1$$(we got two subs in x and 1 in y. cause then we get I start as usually draw it :) https://www.physicsforums.com/attachments/739 then we got the area of sub square $$A=1$$ so we got these point $$f(1,2)+f(2,2)$$
and we get result $$-9$$. Now afterwards I type I notice we can't use the rule right? $$x-3y^2=z$$ and for any of our point we will get $$z<0$$ and then we can't use defination. but are my two above correct?Regards,

Hi Petrus, :)

You have used the midpoint rule incorrectly in the second part where you approximate the integral using two rectangles. Note that you should select points which are midpoints of the two rectangles. Then the integral approximates to,

\[\iint_R(x-3y^2)\,dA\approx f\left(\frac{1}{2},\,\frac{3}{2}\right)+f\left( \frac{3}{2}\,,\frac{3}{2}\right)\]

I don't exactly understand what you meant by,

Now afterwards I type I notice we can't use the rule right? $$x-3y^2=z$$ and for any of our point we will get $$z<0$$ and then we can't use defination.

The integrand can take negative or positive values in the region which it's integrated. There is no problem with that. You might want to check how double integrals are formulated.

 

[Petrus]

Hello,
Yes I understand that. It's because it is continuous function if I got it right, Indeed it says their is for some discontiuos functions but our teacher told us to ignore that cause we will read about it later.

I want to citat from my book
"If $$f(x,y) \geq 0$$, then the volume V of the solid that lies above the rectangle R and below the surface $$z=f(x,y)$$ is
$$V=\int\int_R f(x,y) dA$$"