MHB Double integrals over general regions.

[Petrus]

Hello MHB,
Exemple 3: "Evaluate $$\int\int_D xy dA$$, where D is the region bounded by the line $$y=x-1$$ and parabola $$y^2=2x+6$$"
They say I region is more complicated (the x) so we choose y. so if we equal them we get $$x_1=-1$$ and $$x_2=5$$
then as they said it's more complicated if we work with x limit so we make them to y limit. and we got $$y=\sqrt{2x+6}$$ and $$y=x-1$$ the y limit shall be $$y_1=4$$ and $$y_2=-2$$ but if you put 5 in $$y=\sqrt{2x+6}$$ we get $$y=\pm 4$$ 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 $$-4$$
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 $$y=\sqrt{2x+6}$$ we get $$y=\pm 4$$ so how does this 'exactly' work.

No , this is not correct .

 

[Petrus]

No , this is not correct .

Hello Zaid
$$y=\sqrt{16}$$ what do you mean

Regards,

 

[alyafey22]

$$\sqrt{16} \neq \pm 4$$

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

 

[Petrus]

$$\sqrt{16} \neq \pm 4$$

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

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