Question

Double integration

Answer 1

for \[\int\limits_{y=0}^{\Pi} \int\limits_{x=0}^{\Pi} \cos (x-y) dx dy\] i'm supposed to get 4 but i'm getting 0 so obviously, the way i did it was not good (i integrated in terms of x, then y and substituted the limits in). how are you supposed to integrate this??

Answer 2

if you're struggling with the idea of holding things constant

use \(\cos (x-y) = \cos x \cos y + \sin x \sin y\)

Answer 3

\[\Large \int_{x=0}^{x=\pi}\cos(x-y)dx\]
Let u = x-y. Derive both sides with respect to x to get du/dx = 1 which turns into du = dx. Treat y as a constant (not as a function)
So we'd have
\[\Large \int_{x=0}^{x=\pi}\cos(x-y)dx=\int_{u=-y}^{u=\pi-y}\cos(u)du\]
The limits u = -y and u = pi-y were found by plugging x = 0 and x = pi into u = x-y
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Now we integrate
\[\Large \int_{u=-y}^{u=\pi-y}\cos(u)du = \sin(u)+C\Big]_{u=-y}^{u=\pi-y}\]
\[\Large \int_{u=-y}^{u=\pi-y}\cos(u)du = (\sin(\pi-y)+C)-(\sin(-y)+C)\]
\[\Large \int_{u=-y}^{u=\pi-y}\cos(u)du = \sin(\pi-y)-\sin(-y)\]

Answer 4

From there you integrate that with respect to y (limits y = 0 to y = pi)

Answer 5

that only works with a Jacobian, i think

but it's a great idea

Answer 6

\(u = x - y, \ v = x + y\)??

Answer 7

@IrishBoy123 hmm i'll try doing that right away

Answer 8

@IrishBoy123 I did get to the part with sin(pi-y)-sin(-y) and integrating it gives me 0

Answer 9

@dramaqueen1201 I'm getting 4 when I integrate sin(pi-y)-sin(-y) from y = 0 to y = pi.

Can I see how you're getting 0?

Answer 10

it might help to use the identities

sin(pi-y) = sin(y)
sin(-y) = -sin(y)

Answer 11

which we could have done right at the outset!!!

Answer 12

@jim_thompson5910 well, i wasn't sure how to integrate sin(pi-y) so i tried it on an online calculator and it simplified it to sin y and -sin(-y) is just sin y too, right? so i got \[\int\limits_{y=o}^{\Pi} (\sin y + \sin y)= \int\limits_{y=o}^{\Pi} 2\sin y\] which gives me cos y?

Answer 13

*-cos y

Answer 14

sin(pi-y) = sin(pi)cos(y)-cos(pi)sin(y)
sin(pi-y) = 0*cos(y)-(-1)*sin(y)
sin(pi-y) = 0+sin(y)
sin(pi-y) = sin(y)

Answer 15

when you integrate, you should get -2cos(y). Now evaluate at the limits

Answer 16

@jim_thompson5910 oh... i'm getting 4 now! thanks a ton!

Answer 17

you're welcome