If you mean ##\theta = \frac \pi 4##, then yes, that is a boundary, but ##\theta = \frac {3\pi} 4## isn't a boundary.Wi_N said:Homework Statement
Problem is part of a double integral. but my boundries are:
1<=x^2 + y^2 <=9 so between 2 circles with r1=1 and r2=3
and x<=y and y<=sqrt(3x)
the first boundry is obviously pi/4 and/or 3pi/4
Nor do we, since you haven't told us what you're integrating or shown how you got your result.Wi_N said:the answer is pi/3 and i have no idea how u get that.
Are you assuming that the line boundary and the quadratic boundary intersect on the outer circle?Wi_N said:u obviously have to switch to polar coordinates but x=rcost y=rsint have no resulted in anything.
Have you drawn a picture including ##y =\sqrt 3 x##? What is its slope? Angle of inclination? That will tell you how to get ##\theta## for it.Wi_N said:Homework Statement
Problem is part of a double integral. but my boundries are:
1<=x^2 + y^2 <=9 so between 2 circles with r1=1 and r2=3
and x<=y and y<=sqrt(3x)
the first boundry is obviously pi/4 and/or 3pi/4
the answer is pi/3 and i have no idea how u get that.
Wi_N said:y<=sqrt(3x)
Per the OP, it is ##y \le \sqrt{3x}##, not ##y \le \sqrt 3x##.LCKurtz said:Have you drawn a picture including ##y =\sqrt 3 x##?
I'm guessing that the OP mistyped the problem, especially in light of what he says the answer is supposed to be.Mark44 said:Per the OP, it is ##y \le \sqrt{3x}##, not ##y \le \sqrt 3x##.
Certainly within the realm of possibility.LCKurtz said:I'm guessing that the OP mistyped the problem, especially in light of what he says the answer is supposed to be.
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