The cylinder is projected in a circle on the area Oxy. Use the polar coordinate system:
x = p cos ϕ
y = p sin ϕ
|J| = p
0 <= ϕ <= 2 π
Consider:
x2 + y2 = 3
p2 = 3
p = "\\sqrt{3}"
0 <= p <= "\\sqrt{3}"
"\\iint ydS = \\iint ydxdy = \\int{_0^{2\u03c0}} d\u03d5 \\int{_0^{ \\sqrt{3} }}p\u2219 p sin \u03d5 dp = \n\n\\intop{_0^{2\u03c0}} sin \u03d5\u2219" (p3/3)"\\mid{_0^{ \\sqrt{3}}}d\u03d5 = \n\n\\int{_0^{2\u03c0}} sin \u03d5 \u2219 (3 \\sqrt{3}\/3) d\u03d5 = \\sqrt{3}\u2219 \\intop{_0^{2\u03c0}}sin \u03d5 d\u03d5 = \\sqrt{3} \u2219 (-cos\u03d5 )\n\n\\mid{_0^{2\u03c0}} = - \\sqrt{3}\u2219 (1-1) = 0"
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