Use a Reimann sum to estimate the value

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MarkSA

Junior Member
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Sep 8, 2007
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243
Q:
If R=[-6,18] x [0,12], use a Riemann sum with m=4, n=2 to estimate the value of double integral on R of: (y^2 - 2x^2)dA. Take the sample points to be the upper left corners of the subrectangles.

I've solved that ?x=6, and ?y=6, and so ?A=36. I'm a little confused about how x[sub:26qztzfg]i[/sub:26qztzfg] and y[sub:26qztzfg]j[/sub:26qztzfg] are derived though in the equation:
? ? f(xi,yj)?A. I'm thinking that x[sub:26qztzfg]i[/sub:26qztzfg] = -6 + 6i and y[sub:26qztzfg]j[/sub:26qztzfg] = 6j, but the numbers don't seem to be working out correctly... I ended up with something in the range of V= -99000 when I solved it all out.

When I solved it I used:
V = ?A[f(x[sub:26qztzfg]1[/sub:26qztzfg],y[sub:26qztzfg]1[/sub:26qztzfg])+f(x[sub:26qztzfg]1[/sub:26qztzfg],y[sub:26qztzfg]2[/sub:26qztzfg])+f(x[sub:26qztzfg]2[/sub:26qztzfg],y[sub:26qztzfg]1[/sub:26qztzfg])+f(x[sub:26qztzfg]2[/sub:26qztzfg],y[sub:26qztzfg]2[/sub:26qztzfg]),etc...)
with x[sub:26qztzfg]1[/sub:26qztzfg]=0, y[sub:26qztzfg]1[/sub:26qztzfg]=6, x[sub:26qztzfg]2[/sub:26qztzfg]=6, y[sub:26qztzfg]2[/sub:26qztzfg]=12, etc.

Any ideas what is wrong?
 
MarkSA said:
Q:
If R=[-6,18] x [0,12], use a Riemann sum with m=4, n=2 to estimate the value of double integral on R of: (y^2 - 2x^2)dA. Take the sample points to be the upper left corners of the subrectangles.

I've solved that ?x=6, and ?y=6, and so ?A=36. I'm a little confused about how x[sub:31xbmz00]i[/sub:31xbmz00] and y[sub:31xbmz00]j[/sub:31xbmz00] are derived though in the equation:
? ? f(xi,yj)?A. I'm thinking that x[sub:31xbmz00]i[/sub:31xbmz00] = -6 + 6i and y[sub:31xbmz00]j[/sub:31xbmz00] = 6j, but the numbers don't seem to be working out correctly... I ended up with something in the range of V= -99000 when I solved it all out.

When I solved it I used:
V = ?A[f(x[sub:31xbmz00]1[/sub:31xbmz00],y[sub:31xbmz00]1[/sub:31xbmz00])+f(x[sub:31xbmz00]1[/sub:31xbmz00],y[sub:31xbmz00]2[/sub:31xbmz00])+f(x[sub:31xbmz00]2[/sub:31xbmz00],y[sub:31xbmz00]1[/sub:31xbmz00])+f(x[sub:31xbmz00]2[/sub:31xbmz00],y[sub:31xbmz00]2[/sub:31xbmz00]),etc...)
with x[sub:31xbmz00]1[/sub:31xbmz00]=0, y[sub:31xbmz00]1[/sub:31xbmz00]=6, x[sub:31xbmz00]2[/sub:31xbmz00]=6, y[sub:31xbmz00]2[/sub:31xbmz00]=12, etc.

Any ideas what is wrong?
Click to expand...

If you are told to use 'upper-left' corners, then you must use minimum x, maximum y.

Your x-values would be -6, 0,6,12, and
your y-values would be 6,12.

Your sum would be 36( f(-6,6 ) + f(-6,12 ) +f(0, 6) +f(0,12 ) +f(6,6 ) +f(6,12 ) +f(12, 6) +f(12,12 ) )

I got a value of - 144 for the summation. (times 36, of course)
 
Thanks Paul. So for upper-left corners, the summations basically start with x[sub:3dmk1eo4]0[/sub:3dmk1eo4] and y[sub:3dmk1eo4]1[/sub:3dmk1eo4]?

If this problem was the same, but using upper-right corners, would I then use x[sub:3dmk1eo4]1[/sub:3dmk1eo4] and y[sub:3dmk1eo4]1[/sub:3dmk1eo4]?
 
MarkSA said:
Thanks Paul. So for upper-left corners, the summations basically start with x[sub:345ecph0]0[/sub:345ecph0] and y[sub:345ecph0]1[/sub:345ecph0]?

If this problem was the same, but using upper-right corners, would I then use x[sub:345ecph0]1[/sub:345ecph0] and y[sub:345ecph0]1[/sub:345ecph0]?
Click to expand...

Yes, that seems right.
 
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