Question

Anybody online to help.

Use transformation

Answer 1

\[x=[(u-v)]/3~~~~~~~y=[2u+v]/3\] to evaluate Exactly \[\int\limits_{}^{}~\int\limits_{}^{D}~~e^{x+y}~~~dA\]

Answer 2

\[D{(x,y)|~~1<=~~x+y~~<=4~~~~-4<=~~y=2x<=1}\]

Answer 3

we have to make a variables change first

Answer 4

for x or y

Answer 5

the second part is -4<= y-2x<=1

Answer 6

@ganeshie8 yay!!!

Answer 7

from the equations of transformations, we can write this:
\[\begin{gathered}
1 \leqslant u \leqslant 4 \hfill \\
- 4 \leqslant v \leqslant 1 \hfill \\
\end{gathered} \]

Answer 8

Nice so you're saying \[u=x+y~~~and ~~~v=y-2x\]

Answer 9

yes!

Answer 10

now, we can rewrite your integral as follows:
\[\large I = \int\limits_1^4 {du} \int\limits_{ - 4}^1 {dv} \left| {J\left( {u,v} \right)} \right|f\left( {u,v} \right)\]

Answer 11

its -4 not 4!

Answer 12

wherein:
\[\large \left| {J\left( {u,v} \right)} \right|\]
is the \(Jacobian\) of transormation

Answer 13

transformation*

Answer 14

here is the formula for \(|J|\):
\[\large \left| J \right| = \det \left( {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}} \\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right) = \det \left( {\begin{array}{*{20}{c}}
{1/3}&{ - 1/3} \\
{2/3}&{1/3}
\end{array}} \right) = ...?\]

Answer 15

please compute such determinant
furthermore, we have:
\[\large {e^{x + y}} = {e^u} = f\left( {u,v} \right)\]

Answer 16

\[|J|=-1/9\]

Answer 17

I got this:
\[\large \left| J \right| = \det \left( {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}} \\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right) = \det \left( {\begin{array}{*{20}{c}}
{1/3}&{ - 1/3} \\
{2/3}&{1/3}
\end{array}} \right) = \frac{1}{9} - \left( { - \frac{2}{9}} \right) = ...?\]

Answer 18

How do you compute f(u,v)

Answer 19

oh my bad. yes

Answer 20

it is simple, since \(x+y=u\), then, after a substitution, I get:
\[\Large {e^{x + y}} = {e^u} = f\left( {u,v} \right)\]

Answer 21

oh Ok.

Answer 22

namely, \(f(u,v)\), is our function after the variables change

Answer 23

so, we ca rewrite your integral as follows:
\[\large I = \int\limits_1^4 {du} \int\limits_{ - 4}^1 {dv} \frac{1}{3}{e^u} = \frac{1}{3}\int\limits_1^4 {du\;} {e^u}\int\limits_{ - 4}^1 {dv} = ...?\]

Answer 24

\[1/3~~\int\limits_{-1}^{4}\int\limits_{-4}^{1}~~~~e^{u}~~dudv\]

Answer 25

that's right!
Please keep in mind that you can factorize such integral

Answer 26

you min split

Answer 27

yes!

Answer 28

like below:
\[\huge \begin{gathered}
I = \int\limits_{ - 1}^4 {du} \int\limits_{ - 4}^1 {dv} \frac{1}{3}{e^u} = \hfill \\
\hfill \\
= \frac{1}{3}\int\limits_{ - 1}^4 {du\;} {e^u}\int\limits_{ - 4}^1 {dv} = ...? \hfill \\
\end{gathered} \]