Flux — Question 4

Question 4

A curved dome has more area without proportionately more flux

A hemisphere of radius R lies above the xy plane. A uniform upward flow $\mathbf v=v_0\hat{\mathbf z}$ crosses its curved surface, oriented outward from the hemisphere. Use polar angle $\theta$ measured from positive z.

1. Integrate the flux through the curved dome and compare it with $v_0$ times the curved area.

2. Include the flat base to make a closed surface and find the total outward flux.

Question 4 - Solution

A curved dome has more area without proportionately more flux

(a) Method and calculation

$$dA=R^2\sin\theta\,d\theta\,d\phi,\quad\mathbf v\cdot\hat{\mathbf n}=v_0\cos\theta.$$$$Q_{\mathrm{dome}}=2\pi R^2v_0\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta=\pi R^2v_0.$$

(b) Interpretation and further calculation

$$A_{\mathrm{dome}}=2\pi R^2,\quad Q_{\mathrm{dome}}=\tfrac12v_0A_{\mathrm{dome}}.$$$$Q_{\mathrm{base}}=-\pi R^2v_0,\quad Q_{\mathrm{closed}}=0.$$The dome passes the same net volume as its projected circular opening.

Physical check

The base normal of the closed hemisphere points downward, so its outward flux is negative.