Figuring out the bounds of an integral

[pakman]

I understand how to solve double integrals, but I am having trouble figuring out what the bounds are. This is a problem we did in class:

Evaluate the integral (assume S = integral)

SS (4 - x^2 - y^2) dxdy where R is the first quadrant sector of the circle x^2 + y^2 = 4 between the lines y = 0 and x = 0. For reference there is supposed to be an R under the double integral.

The teacher came up with these polar coordinates

0 <= r <= 2 and 0 <= theta <= pi/4

I understand how he got the bounds for r but what about for theta? Thanks in advance

 

[soroban]

Hello, pakman!

I understand how to solve double integrals,
but I am having trouble figuring out what the bounds are.
This is a problem we did in class:

Evaluate the integral: \(\displaystyle \L\:\int_{\;R}\int(4\,-\,x^2\,-\,y^2)\,dx\,dy\)
where \(\displaystyle R\) is the first quadrant sector of the circle \(\displaystyle x^2\,+\,y^2\:=\:4\)

The teacher came up with these polar limits:
. . \(\displaystyle 0\,\leq\,r\,\leq\,2\,\) and \(\displaystyle \,0\,\leq\,\theta\,\leq\,\frac{\pi}{4}\)

I understand how he got the bounds for \(\displaystyle r\), but what about for \(\displaystyle \theta\) ?

I agree with you . . . Your teacher is wrong.

The limits for \(\displaystyle \theta\) are: \(\displaystyle \,\left[0,\,\frac{\pi}{2}\right]\)

 

[galactus]

I agree with Japanese for abacus . I thought I wasn't seeing something. It's that portion of the circle in the first quadrant. A quarter circle. Therefore, 0 to pi/2.