Question

will award medal
Let R be the rectangle with vertices (0,0), (2,0), (2,2), and (0,2) and let f(x,y) = \sqrt{xy}.

(a) Find reasonable upper and lower bounds for \int_{R}f\,dA without subdividing R.

Answer 1

@Zarkon youre smart as hell, anyway you could help me out? to at least start this problem

Answer 2

depends on what they mean by reasonable.

I would take \(\sqrt{0\cdot0}\) and \(\sqrt{2\cdot2}\) as lower and upper bounds for \(f\)

then multiply by the area of \(R\)

Answer 3

\[0\le \sqrt{xy}\le2\]

so \[\iint\limits_{R}0dA\le\iint\limits_{R}\sqrt{xy}dA\le\iint\limits_{R}2dA\]

Answer 4

\[0\times Area(R)\le\iint\limits_{R}\sqrt{xy}dA\le2\times Area(R)\]

\[0\le\iint\limits_{R}\sqrt{xy}dA\le2\times4=8\]

Answer 5

so 8 would be the upper bound

Answer 6

yes...though it is not a great upper bound

Answer 7

and zero is the lower bound

Answer 8

yes

Answer 9

the true value of the integral is 32/9 so I guess 0 and 8 aren't terrible

Answer 10

the webwork likes em. and for the second part

Answer 11

for making a table what would i do to set that up?

Answer 12

so would my delta x and delta y be 2?

Answer 13

no

Answer 14

take each square at a time. find the max and min of \(\sqrt{xy}\) on each square

Answer 15

would it be this? @Zarkon