If (a,b,c) is normal to a surface, so is (-a,-b,-c). There are two normals to a surface at a given point. For example in the first one you could also have written the plane -x-y-z=0. It's the same plane.
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Dick said:
If (a,b,c) is normal to a surface, so is (-a,-b,-c). There are two normals to a surface at a given point. For example in the first one you could also have written the plane -x-y-z=0. It's the same plane.
Yeah, but why did it switch it on the second one, and the first one stayed positive?
If I learned anything, it is that there is no random in math...
Um, what do you mean by "switch"? I can't make out the coefficient of y in the equation of plane for the second one. You could use the right-hand screw rule along with r_u X r_v to see which direction the normal vector points. It all depends on how you parametrise it. If you had done u=y, v=x for the second one, it would have been the same as the normal vector read off the equation of the plane.
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$
Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$
The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$
Plugging identity above into ##(1)## with ##u...