Problem with the region of an integral

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Homework Statement
look at the image
Relevant Equations
Stocke theorem!
Greetings!

The exercice ask to calculate the circuitation of the the vector field F on the border of the set omega
I do understand the solution very well
my problem is the region!
I m used to work with a region delimitated clearly by two intersecting function here the upper one stop a y=3 and doesn´t intersect with y=0.

thank you!
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Amaelle said:
I m used to work with a region delimitated clearly by two intersecting function here the upper one stop a y=3 and doesn´t intersect with y=0.
No, the upper function doesn't stop at y = 3. The quarter circle does, but the function is given by ##y = 3 + \sqrt{9 - x^2}##.
The region ##\Omega## is (obviously) the portion in blue. The inequality ##0 \le y \le 3 + \sqrt{9 - x^2}## includes all those y-values for which ##x \ge 0##, that run from the x-axis up to the quarter circle. For example, if x = 0, y ranges from 0 to 6; if x = 3, y ranges from 0 to 3.
 
Mark44 said:
No, the upper function doesn't stop at y = 3. The quarter circle does, but the function is given by ##y = 3 + \sqrt{9 - x^2}##.
The region ##\Omega## is (obviously) the portion in blue. The inequality ##0 \le y \le 3 + \sqrt{9 - x^2}## includes all those y-values for which ##x \ge 0##, that run from the x-axis up to the quarter circle. For example, if x = 0, y ranges from 0 to 6; if x = 3, y ranges from 0 to 3.
Thanks a million!
 
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