Question

find the volume under the graph of the function f(x,y)=6x^2y over the region of the triangle with verticies of (1,3)(3,9)(3,3)

Answer 1

fun double integral problem

Answer 2

(1,3)(3,9)(3,3)

derive function

Answer 3

right..i'm there

Answer 4

i get y = 3x and y = 3....and the third one is undefined

Answer 5

but this is where i'm lost in setting up the integral i guess

Answer 6

should be the integral of 1 to 9 and the other integral for terms of x is screwing with me

Answer 7

sorry 3 to 9

Answer 8

choose
dx dy

x should go from x=1/3y to x=3
y should go from 3 to 9

Answer 9

can u explain how u got the x values?

Answer 10

i'm confused

Answer 11

right...the x = 1/3y...how so?

Answer 12

since the slope is 0....y = 3

Answer 13

so is it just x = 1/3y?

Answer 14

and an assumed x value?

Answer 15

y=3x (as you found out)
solve for x

y/3= x

Answer 16

ohhhh

Answer 17

ur using the same one okay

Answer 18

ty sooo much

Answer 19

\[\int _3^9\int _{\frac{y}{3}}^31dxdy=\int _1^3\int _3^{3x}1dydx\]

Answer 20

what's more important is that \[\int _3^9\int _{\frac{y}{3}}^36x^2y\,dxdy=\int _1^3\int _3^{3x}6x^2y\,dydx\]