PX275 - D4c - total vector surface area

to evaluate the vector area:

\vec{S} = \iint d \vec{s}

the total vector surface area of the flat disc

{\vec{S}}_{1} = \iint d \vec{s} = \iint \hat{n} d A

d \vec{s} = d A \hat{k} = ρ d ϕ d ρ \hat{k}

\begin{aligned} {\vec{S}}_{1} & = \int_{ρ = 0}^{a} \int_{ϕ = 0}^{2 π} ρ d ϕ d ρ \hat{k} \\ = {[2 π \frac{ρ^{2}}{2} \hat{k}]}_{0}^{a} \\ {\vec{S}}_{1} & = π a^{2} \hat{k} \end{aligned}

the total vector surface area of the dome of the hemisphere

{\vec{S}}_{2} = \iint d \vec{s} = \iint \hat{n} d A

considering the $\hat{i}, \hat{j}, \hat{k}$ components individually for the hemisphere

the $\hat{k}$ -component

\begin{aligned} {\vec{S}}_{2 (k)} & = \int d A \cos θ \hat{k} \\ = \iint r^{2} \sin θ d θ d ϕ \cos θ \hat{k} \\ = \int_{0}^{\frac{π}{2}} \int_{0}^{2 π} a^{2} \sin θ \cos θ d ϕ d θ \hat{k} \\ = 2 π a^{2} \hat{k} \int_{0}^{\frac{π}{2}} \sin θ \cos θ d θ \\ = 2 π a^{2} {[\frac{\sin^{2} θ}{2}]}_{0}^{\frac{π}{2}} \hat{k} \\ = π a^{2} \hat{k} \end{aligned}

note: $\cos θ$ because this is the projection of $\hat{n}$ on $\hat{k}$

the $\hat{i}$ -component

the $\hat{i}$ -component of ${\hat{e}}_{r}$ is $\sin θ \cos ϕ$

\begin{aligned} {\vec{S}}_{2 (i)} & = \iint d A \sin θ \cos ϕ \hat{i} \\ = \int_{0}^{\frac{π}{2}} \int_{0}^{2 π} r^{2} \sin θ d θ d ϕ \sin θ \cos ϕ \hat{i} \\ = \int_{0}^{\frac{π}{2}} a^{2} \sin^{2} θ d θ \int_{0}^{2 π} \cos ϕ d ϕ \hat{i} \\ = 0 \end{aligned}

the $\hat{j}$ -component

it can be deduced from symmetry that $S_{2 (i)}, S_{2 (j)} = 0$

\begin{aligned} {\vec{S}}_{2 (j)} & = \iint r^{2} \sin θ d θ d ϕ \sin θ \sin ϕ \hat{j} \\ = \int_{0}^{\frac{π}{2}} a^{2} \sin^{2} θ d θ \int_{0}^{2 π} \sin ϕ d ϕ \hat{j} \\ = 0 \end{aligned}

hence, the total vector surface area of the hemisphere is:

{\vec{S}}_{2} = π a^{2} \hat{k}