Exercise · ZW-DOUBLE-INTEGRAL-010 · ★★★ Advanced
Interpret \(\iint_D \sqrt{1-x^2-y^2}\,dA\), with \(D:\ x^2+y^2\le1\), as a volume and compute it.
Choose one option.
Show answer
- C. \(\tfrac{2\pi}{3}\)
Explanation
The graph of \(z=\sqrt{1-x^2-y^2}\) is the upper hemisphere of radius \(1\), and the integral is the volume between it and the disk \(D\): \(\tfrac12\cdot\tfrac43\pi=\tfrac{2\pi}{3}\). In polar form: \(2\pi\int_0^1\sqrt{1-r^2}\,r\,dr=2\pi\cdot\tfrac13\). The whole sphere would give \(\tfrac{4\pi}{3}\).
Review the concept: Double integral
The double integral of a function \(f\) over a region \(D\) of the plane is the limit of the sums of the values of \(f\) times the area \(\Delta A\) of ever finer cells covering \(D\).
\[\iint_D f(x,y)\,dA=\lim_{\max\Delta A_i\to0}\sum_i f(x_i^*,y_i^*)\,\Delta A_i\]
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