Question

Evaluate the integral ∫R2xydA, where R is the triangle 2x+y≤2, x≥0, y≥0

Answer 1

hey mr. math

Answer 2

u were there earlier with james right?

Answer 3

Yes.

Answer 4

okay...we came up with the integral...but it seems like my bounds are incorrect

Answer 5

What were your bounds?

Answer 6

we came up with x from 0 to 1...and y from 0 to -2x+2

Answer 7

2xy dydx

Answer 8

This is correct.

Answer 9

This is correct.

Answer 10

how come i can't run that through wolfram?

Answer 11

right we got all that...and looks the same...i try to run it through wolfram to get an exact answer and it doesn't work

Answer 12

It should look like\[\int\limits_0^1 \int\limits_0^{2-2x} 2xydydx\]

Answer 13

ohhh

Answer 14

lemme try that

Answer 15

unreal...i kept getting -1/3....so i ran it through wolfram cause i thought there was an error....the answer is 1/3

Answer 16

\[=\int\limits_0^1xy^2|_0^{2-2x}dx=\int\limits_0^1x(2-2x)^2=4\int\limits_0^1x^3-2x^2+x\]
\[=(x^4-\frac{8}{3}x^3+2x^2)|_0^1=1-\frac{8}{3}+2=\frac{1}{3}.\]

Answer 17

ya i got it now...thx soo much for helping

Answer 18

now try the order of integration reversed