0
$\begingroup$

Verify Stokes theorem when $F=(y^2 + z^2 - x^2)i+ (z^2 + x^2 - y^2)j+(x^2 + y^2 -z^2)k$ and $s$ is portion of surface $x^2+y^2-2ax+az=0$ above the plane $z=0$

In the following question what would be the polar coordinates for x and y?I have already tried taking the usual $x=r\cos q$ and $y=r \sin q$ but then the area and the line integral is not equal and yes iam not pretty good at mathematics.

1 Answers 1

0

Hint. Note that $x^2+y^2-2ax+az=0$ can be written as: $$\left(x-a\right)^2+y^2+az=a^2$$ So for $z=0$, you get: $$\left(x-a\right)^2+y^2=a^2$$ This is a circle in the $xy$-plane, centered in ($a,0$) and with radius $a$. For the line integral, you could parametrize as: $$x = a+a\cos t \;,\; y = a \sin t \quad;\quad 0 \le t \le 2\pi$$ For the surface integral, the projection of the surface $s$ onto the $xy$-plane will be the interior of this circle (disc) so for the double integral, you can take: $$x = a+r\cos t \;,\; y = r \sin t \quad;\quad 0 \le t \le 2\pi \; ,\; 0 \le r \le a$$

  • 0
    Because the x-coordinate of the circle's center is not 0 but a. You can plug it in and expand the square to see that it works. – 2017-01-13
  • 0
    So if y coordinate of circle would have been b then y would have been taken as b+asint – 2017-01-13
  • 0
    Yes, indeed...! – 2017-01-13