Calculus 3 quick question: What am I doing wrong here ? ∫ ∫ x^2+y^2 dxdy over the region D={(x,y) belonging to R^2 | 1

Question

Calculus 3 quick question: What am I doing wrong here ?

∫ ∫ x^2+y^2 dxdy over the region D={(x,y) belonging to R^2 | 1 <= x^2+y^2<=2*x, y<=0}

In polar coordinates we have 1<=r^2<=2*r*cos(theta). Because r is positive, we could say r<=2*cos(theta) on the right side.

Now, y>=0 that means 0<=theta<=pi. As we know, cos(theta) is negative between (pi/2,pi), which will make r<= something negative which doesn't make sense at all.

mathmale · Answer

While I believe I can decipher your region R, it'd help me, and perhaps help you, if you'd please DRAW that region using the Draw utility (below) for confirmation.

mathmale · Answer

Please confirm that I've interpreted your typing correctly:
$$1\le (x ^{2}+y ^{2}\le 2x;y \le0$$

mathmale · Answer

If so, you may want to try moving that 2x term to the middle of this inequality and then completing the square for y:  $$1\le x ^{2}+y ^{2}-2y \le0;y \le0$$

mathmale · Answer

... which, if I've done the work correctly, boils down to 
$$1\le x ^{2}+(y-1)^{2}\le 1$$

mathmale · Answer

which may or may not make sense.  Sorry I can't stick with you right now; hope this info helps you determine that region R so that y ou can identify the proper limits of integration.  Good luck!

ganeshie8 · Answer

|dw:1390146570417:dw|