MHB How Do You Solve This Complex Double Integral with Given Curves?

[lfdahl]

Evaluate the double integral:
\[I = \int \int _R\frac{1}{(1+x^2)y}dxdy\]
- where $R$ is the region in the upper half plane between the two curves:

$2x^4+y^4+ y = 2$ and $x^4 + 8y^4+y = 1$.

 

[Albert1]

Evaluate the double integral:
\[I = \int \int _R\frac{1}{(1+x^2)y}dxdy\]
- where $R$ is the region in the upper half plane between the two curves:

$2x^4+y^4+ y = 2$ and $x^4 + 8y^4+y = 1$.

please give a hint how to find the region of $R$
for $ y $ is hard to express in $x$

Spoiler

=$2ln(2)\times [tan^{-1}(1-y-8y^4 )^\dfrac{1}{4} - tan^{-1}(1. - \dfrac{y}{2}-\dfrac{y^4}{2})^\dfrac{1}{4}]$
or $\dfrac {\pi}{2} ln(y)$

 

[lfdahl]

Spoiler

=$2ln(2)\times [tan^{-1}(1-y-8y^4 )^\dfrac{1}{4} - tan^{-1}(1. - \dfrac{y}{2}-\dfrac{y^4}{2})^\dfrac{1}{4}]$
correct ?

Spoiler

No. :(. Are you sure about your integration limits?

 

[Albert1]

Spoiler

No. :(. Are you sure about your integration limits?

hint for range of y

Spoiler

$x$ from -1 to 1
but how to express $y$ in $x$
the range of y?

$$\int_{y_1}^{y_2}\dfrac{dy}{y}\int_{-1}^{1}\dfrac{dx}{1+x^2}\\
=\dfrac{\pi}{2}ln(\dfrac{y_2}{y_1})$$
$$=2\int_{0}^{1}\dfrac{dx}{1+x^2}\int_{y_1}^{y_2}\dfrac{dy}{y}$$
$$=2\int_{0}^{1}\dfrac{ln(\dfrac{y_2}{y_1})}{1+x^2}dx$$
$=\dfrac {\pi}{2}\times ln(2)$
from Ifdahl 's hint $y_2=2y_1$
for same $x$

 

[lfdahl]

hint for range of y

Spoiler

$x$ from -1 to 1
but how to express $y$ in $x$
the range of y?

$$\int_{y_1}^{y_2}\dfrac{dy}{y}\int_{-1}^{1}\dfrac{dx}{1+x^2}\\
=\dfrac{\pi}{2}ln(\dfrac{y_2}{y_1})$$
$$=2\int_{0}^{1}\dfrac{dx}{1+x^2}\int_{y_1}^{y_2}\dfrac{dy}{y}$$
$$=2\int_{0}^{1}\dfrac{ln(\dfrac{y_2}{y_1})}{1+x^2}dx$$

Spoiler

Your $x$-range is correct.
Can you show, that $y_2 = 2y_1$?
Then you have cracked the nut :)

 

[Albert1]

Spoiler

Your $x$-range is correct.
Can you show, that $y_2 = 2y_1$?
Then you have cracked the nut :)

solution for :$y_2=2y_1$

Spoiler

$y_2^4+y_2=2-2x^4---(2)$
$8y_1^4+y_1=1-x^4---(1)$
$\dfrac {(2)}{(1)}$:
we have $\dfrac {y_2^4+y_2}{8y_1^4+y_1}=2$
so $y_2=2y_1$(for same $x)$
and the answer is $\dfrac {\pi}{2}\times ln(2)$

 

[lfdahl]

solution for :$y_2=2y_1$

Spoiler

$y_2^4+y_2=2-2x^4---(2)$
$8y_1^4+y_1=1-x^4---(1)$
$\dfrac {(2)}{(1)}$:
we have $\dfrac {y_2^4+y_2}{8y_1^4+y_1}=2$
so $y_2=2y_1$(for same $x)$
and the answer is $\dfrac {\pi}{2}\times ln(2)$

Thankyou for your participation, Albert! Well done!