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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

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 cant use lower limit as \(\displaystyle -4\)

What I exactly mean why dont we have our lowest limit as -4 insted of -2

(It may be little confusing, sorry)

Regards,