Calculate $\iint_{S} x^2ds$ over cyliderical surface $x^2+y^2=4$
Method (i): $\iint_{S} x^2 \frac{dxdz}{\hat{j}.\hat{n}} = \iint_{S} x^2 \frac{dxdz}{y/2} \qquad (\because \hat{n}=\frac{x\hat{i}+y\hat{j}}{2})$
$=\iint_{S} 2x^2 \frac{dxdz}{\sqrt{4-x^2}}\qquad (taking \; x=2cos\theta, dx=-2sin\theta d\theta \;$ and integrating in positive quadrant$)$
$=4*\iint_{S/4} -8cos^2\theta d\theta dz$
Method (ii): $\iint_{S} x^2 2d\theta dz \qquad (taking \; x=2cos\theta,$ and integrating$)$
$=\iint_{S} 8cos^2\theta d\theta dz$
Doubt: Why I get minus sign difference here. I'm not able to visualise the difference. Please help.