Double integrals over general regions.

[Petrus]

Hello MHB,
Exemple 3: "Evaluate \(\displaystyle \int\int_D xy dA\), where D is the region bounded by the line \(\displaystyle y=x-1\) and parabola \(\displaystyle y^2=2x+6\)"
They say I region is more complicated (the x) so we choose y. so if we equal them we get \(\displaystyle x_1=-1\) and \(\displaystyle x_2=5\)
then as they said it's more complicated if we work with x limit so we make them to y limit. and we got \(\displaystyle y=\sqrt{2x+6}\) and \(\displaystyle y=x-1\) the y limit shall be \(\displaystyle y_1=4\) and \(\displaystyle y_2=-2\) but if you put 5 in \(\displaystyle y=\sqrt{2x+6}\) we get \(\displaystyle y=\pm 4\) so how does this 'exactly' work. Shall I check the value on both function? I reread the chapter and try think and think but never come up with an answer why I can't use lower limit as \(\displaystyle -4\)
What I exactly mean why don't we have our lowest limit as -4 insted of -2
(It may be little confusing, sorry)Regards,

 

[alyafey22]

but if you put 5 in \(\displaystyle y=\sqrt{2x+6}\) we get \(\displaystyle y=\pm 4\) so how does this 'exactly' work.

No , this is not correct .

 

[Petrus]

No , this is not correct .

Hello Zaid
\(\displaystyle y=\sqrt{16}\) what do you mean

Regards,

 

[alyafey22]

\(\displaystyle \sqrt{16} \neq \pm 4\)

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For this example it is better to separate the integral , we cannot define \(\displaystyle y=\sqrt{2x+6}\) because $y$ takes negative values.

 

[Petrus]

\(\displaystyle \sqrt{16} \neq \pm 4\)

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For this example it is better to separate the integral , we cannot define \(\displaystyle y=\sqrt{2x+6}\) because $y$ takes negative values.

The point is we can get y value two way. \(\displaystyle y=x-1\) and \(\displaystyle y=\sqrt{2x+6}\) in the first one if we put 5 we get positive 4 so I can just put in first one :)?