How do I calculate this? How do I find dS? What are the boundaries of the integral? Thanks!
2026-03-25 04:40:52.1774413652
Calculate the surface integral of the cylinder that cuts the cone
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HINTS
Note that the equation of the cylinder is
$$x^2+y^2=2x \implies x^2-2x+1+y^2=1\implies (x-1)^2+y^2=1$$
thus it is generate by a circle with radius $1$ and center at $(1,0)$.
For a fixed $z$, the intersection between the culynder ant the cone is given by
$$x^2+y^2=2x=z^2\implies x=\frac{z^2}2 \quad y=\sqrt{z^2-\frac{z^4}4}$$
The integrand function becomes
$$x^4-y^4+y^2z^2-z^2x^2+1=x^4-y^4+(y^2-x^2)(y^2+x^2)+1=1$$