Respuesta :
Solution :- To find the vector area of hemispherical bowl of radius r, the vector area is given by the integral
[tex]\vec{a}=\int_{s} d\vec{a}[/tex] for a surface S
The area for a hemisphere is given by
[tex]d\vec{a}=R^{2} \sin\theta d\theta d \phi \hat{r}[/tex]
[tex]\vec{a} = R^{2}\int_{0}^{\pi /2} sin\theta d\theta \cdot \int_{0}^{2\pi}d\phi \hat{r}[/tex]
by using}
[tex]\hat{r} = sin\theta cos \phi \hat{i}+sin\theta sin\phi \hat{j}+cos\theta \hat{k}[/tex]
[tex]\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} ][/tex]
[tex]=\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[/tex]
[tex]=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}[/tex]
The vector area of the hemispherical bowl of radius r is equal to [tex]\pi R^2\hat k[/tex].
What is the vector area?
The vector of a flat surface whose magnitude is the figure's area and whose direction is perpendicular to the figure's plane
The vector area is given by the formula,
[tex]\vec a= \int_s d\vec a[/tex]
And the area of the hemisphere is given by the formula,
[tex]d\vec a= R^2\rm sin\theta d\theta d\phi \hat{r}[/tex]
[tex]\vec a = R^2 \int_0^{\frac{\pi}{2}}Sin\theta d\theta\ \cdot\ \int_0^{\frac{\pi}{2}}d\phi \hat r[/tex]
Also, the radius of the hemisphere can be written as,
[tex]\hat r= sin\theta cos\phi \hat i+ sin\theta cos\phi \hat j+ sin\theta cos\phi \hat k[/tex]
Therefore, the surface area of the hemisphere can be written as,
[tex]\vec a = R^2 \int_0^{\frac{\pi}{2}}Sin\theta d\theta\ \cdot\ \int_0^{\frac{\pi}{2}}d\phi (sin\theta cos\phi \hat i+ sin\theta cos\phi \hat j+ sin\theta cos\phi \hat k)[/tex] [tex]=\hat iR^2\int_0^{\frac{\pi}{2}} sin^2\theta d \phi \int_0^{\frac{\pi}{2}}cos\phi d\phi+\hat j R^2 \int_0^{\frac{\pi}{2}} sin\theta d \theta \int_0^{\frac{\pi}{2}}sin\phi d\phi+\hat k R^2\int_0^{\frac{\pi}{2}} sin\theta cos \theta d \theta\int_0^{\frac{\pi}{2}} d\phi[/tex]
As the value of Cosθ=t, -sinθdθ=dt, therefore,
[tex]= 0 + 0 + 2\pi R^2\hat k\int_0^1(-1)tdt[/tex]
[tex]=\pi R^2 \hat k[/tex]
Hence, the vector area of the hemispherical bowl of radius r is equal to [tex]\pi R^2\hat k[/tex].
Learn more about Vector:
https://brainly.com/question/16000644
