Question

∫∫(5−(4/3)y)/(1+x)^2dxdy over the area that is enclosed by the curves 4-4/3*y-x=0 and 1/3*y^2=x.

Answer 1

Would this be the correct volume?
\[\int\limits_{0}^{4/3}(\int\limits_{-\sqrt(3x)}^{\sqrt(3x)}(5-\frac{4}{3}y)dy)dx+\int\limits_{4/3}^{12}(\int\limits_{-\sqrt(3x)}^{3-\frac{3}{4}x}(5-\frac{4}{3}y)dy)dx\]

Answer 2

dxdy will make it easy

Answer 3

What do you mean?

Answer 4

switch the order

Answer 5

make it dxdy, instead of dydx

Answer 6

Won't that fetch me an end result with a bunch of x?

Answer 7

Oh crap, that's not the right integrands above; they're missing *1/(1+x)^2

Answer 8

before we talk about that, let me point out something

Answer 9

yeah integrand will not change

Answer 10

oh u forgot putting that.. okay i got u :)

Answer 11

consider switching the order... that makes the bounds very easy and gives u oly one integral to evaluate

Answer 12

yes, but won't I just end up with an expression filled with x?

Answer 13

If I switch the iterated integration

Answer 14

dxdy gives u below :
outer integral : -6 to 2
inner integral : y^2/3 to 4-4/3*y

Answer 15

you're saying integrating is a pain ?

Answer 16

\(\large \int \limits_{-6}^2 ~~\int \limits_{y^2/3}^{4-4/3y} \frac{5-4/3y}{(1+x)^2}dxdy\)

Answer 17

ooooooh; you mean I could just "rotate" the whole plot?

Answer 18

yes :)

Answer 19

Well, dang! Thanks a lot; I'll try that out. :)

Answer 20

np... good luck !