Evaluate the surface integral. $\iint_{S} y^{2} d S$ $S$ is the part of the sphere $x^{2}+y^{2}+z^{2}=1$ that lies above the cone $z=\sqrt{x^{2}+y^{2}}$
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Step 1: Parameterize the unit sphere using spherical coordinates: x = sin(phi) cos(theta), y = sin(phi) sin(theta), z = cos(phi), with 0 ≤ theta ≤ 2π and 0 ≤ phi ≤ π. Show more…
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Evaluate the surface integral. $$ \begin{array}{l}{\iint_{S} y^{2} d S} \\ {S \text { is the part of the sphere } x^{2}+y^{2}+z^{2}=1 \text { that lies above }} \\ {\text { the cone } z=\sqrt{x^{2}+y^{2}}}\end{array} $$
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