Question

Find the vector area of a hemispherical bowl of radius r.

Answer 1

Solution :- To find the vector area of hemispherical bowl of radius r, the vector area is given by the integral

$\vec{a}=\int_{s} d\vec{a}$    for a surface S

The area for a hemisphere is given by

$d\vec{a}=R^{2} \sin\theta d\theta d \phi \hat{r}$

$\vec{a} = R^{2}\int_{0}^{\pi /2} sin\theta d\theta \cdot \int_{0}^{2\pi}d\phi \hat{r}$

by using}

$\hat{r} = sin\theta cos \phi \hat{i}+sin\theta sin\phi \hat{j}+cos\theta \hat{k}$

$\text{Now we have }\\ \vec{a} = R^{2}\int_{0}^{\pi /2} sin\theta d\theta \int_{0}^{2\pi}d\phi [sin\theta cos\phi\hat{i}+sin\theta sin\phi \hat{j}+cos\theta \hat{k} ]$

$=\hat{i}R^{2}\int_{0}^{\pi/2}sin^{2}\theta d\phi \int_{0}^{2\pi } cos\phi d\phi + \hat{j}R^{2}\int_{0}^{\pi /2}sin\theta d\theta \int_{0}^{2 \Pi } sin\phi d\phi+\hat{k}R^{2}\int_{0}^{\pi/2}sin{\theta}cos{\theta}d{\theta}\int_{0}^{2\pi}d\phi$

$=0+0+2{\pi}R^2\hat{k}\int_{0}^{1}(-1)tdt\\ \text {[here} \cos\theta=t\\-sin\theta d\theta=dt]\\= \pi R^2 \hat{k}$