Question

Will award a medal
@ganeshie8

Answer 1

i got 17/2

Answer 2

my new transformation integral was \[1/2\int\limits_{0}^{3}\int\limits_{0}^{2}(-2/3u+7/3v)dudv\]

Answer 3

@IrishBoy123

Answer 4

why 1/2 ?

Answer 5

the Jacobian is 1/3

Answer 6

I got 17/3, is it right?

Answer 7

i got the jacobian as 1/2 but that could be wrong.

Answer 8

with all the denominators as 3, there is no way for Jacobian to be 1/2

Answer 9

1/6+2/6

Answer 10

you should be working

1/9 + 2/9

Answer 11

ohghad 3x3=9

Answer 12

here is my Jacobian:
\[\Large J = \left( {\begin{array}{*{20}{c}}
{1/3}&{1/3} \\
{ - 2/3}&{1/3}
\end{array}} \right),\quad \left| J \right| = \left( {\frac{1}{3} \times \frac{1}{3}} \right) - \left\{ {\frac{1}{3} \times \left( { - \frac{2}{3}} \right)} \right\}\]

Answer 13

\[\Large \begin{gathered}
J = \left( {\begin{array}{*{20}{c}}
{1/3}&{1/3} \\
{ - 2/3}&{1/3}
\end{array}} \right),\quad \hfill \\
\hfill \\
\left| J \right| = \left( {\frac{1}{3} \times \frac{1}{3}} \right) - \left\{ {\frac{1}{3} \times \left( { - \frac{2}{3}} \right)} \right\} \hfill \\
\end{gathered} \]

Answer 14

for this one, is the parameterization x=0,y=y,z=f(y)sin(theta)

Answer 15

@ganeshie8

Answer 16

hint:
I rewrite your equation, as below:
\[\large y = \frac{{{z^2}}}{{25}} + 1\]
so we have: