The problem involves determining the boundaries for a double integral, specifically within the context of polar coordinates. The original poster describes the region defined by two circles and additional linear boundaries.
The discussion is ongoing, with participants exploring various interpretations of the boundaries and questioning the original poster's setup. Some guidance has been offered regarding the need to visualize the problem and consider the slopes of the lines involved.
There are indications that the original poster may have made typographical errors in the problem statement, which could affect the interpretation of the boundaries. The specific function being integrated has not been provided, leaving some ambiguity in the discussion.
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 boundary 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 boundary 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.