Exercise · ZW-SURFACE-INTEGRAL-009 · ★★★ Advanced
The area of the part of the paraboloid \(z=x^2+y^2\) with \(z\le1\) is:
Choose one option.
Show answer
- D. \(\dfrac{\pi}{6}\left(5\sqrt5-1\right)\)
Explanation
\(dS=\sqrt{1+4r^2}\,r\,dr\,d\theta\) with \(0\le r\le1\): \(2\pi\cdot\tfrac1{12}\left(5\sqrt5-1\right)=\tfrac{\pi}{6}\left(5\sqrt5-1\right)\).
Review the concept: Surface integral
The surface integral of a function \(f\) over a surface \(S\), parametrized by \(\mathbf r\) over a region \(D\), adds up the values of \(f\) weighted by the area of each piece of \(S\).
\[\iint_S f\,dS=\iint_D f(\mathbf r(u,v))\,\lVert\mathbf r_u\times\mathbf r_v\rVert\,du\,dv\]
More practice problems: surface integral
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