Question

(a) to find the surface area of the sphere using integration.

(b)To find the centroid of curve surface area of hemisphere.

Short answer

(a). The surface area of the sphere using integration is$4\pi a^2$.

(b). The centroid of curve surface area of hemisphere is$\left(0,0,\frac{a}{2}\right)$ .

Step-by-step solution

Given information

For the sphere $r=a$.

Formula of surface area of sphere

The surface area of the sphere is calculated as,$\mathit{A}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{2}\mathbf{\pi }}{\mathit{a}}^{\mathbf{2}}\mathit{d}\mathit{\varphi }\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{\theta }\mathbf{\right)}\mathit{d}\mathit{\theta }$.

Calculation of the surface area of sphere

(a)

By the formula of sphere calculate the surface area.

$$\begin{gathered} A={\int }_0^{\pi }{\int }_0^{2\pi }{a}^{2}d\succ \sin \left(\theta \right)d\theta \\ =a^2[\succ {]}_0^{2\pi }[-\cos \theta {]}_0^{\pi } \\ =a^2(2\pi -0)(-1)(\cos \pi -\cos 0) \\ =2\pi a^2(-1)(-1-1) \end{gathered}$$

Solve further, to obtain$A=4\pi a^2$

Thus, the surface area of the sphere using integration is$4\pi a^2$ .

Calculation of  z coordinate of the centroid of the curved surface area of hemisphere

(b)

Calculate z coordinates of the centroid as follows:

$$\begin{gathered} \overline{x}=\frac{\int xda}{\int da} \\ =\frac{{\int }_0^{se}{\int }_0^{2\pi }{a}^3\cos \succ d\succ {\sin }^2\theta d\theta }{{\int }_0^{\pi /2}{\int }_0^{2\pi }{a}^{2}d\succ \sin \theta d\theta } \\ =\frac{a^3{\int }_0^{\Omega /2}\frac{(1-\cos 2\theta )}{2}d\theta ·{\int }_0^{2x}\cos \succ d\succ }{a^2[-\cos \theta {]}_0^{\alpha /2}[\succ {]}_0^{2x}} \\ =\frac{\frac{a}{2}{\left(\theta -\frac{\sin 2\theta }{2}\right)}_0^{2/2}·[\sin \succ {]}_0^{2\pi }}{-\left(\cos \frac{\pi }{2}-\cos 0\right)(2\pi -0)} \end{gathered}$$

Solve further as follows:

$$\begin{gathered} \overline{x}=\frac{\frac{a}{2}\left(\frac{x}{2}-\sin \theta -0\right)(\sin 2\pi -0)}{2\pi (-1)(0-1)} \\ \overline{x}=0 \end{gathered}$$

Calculation of   y coordinate of the centroid of the curved surface area of hemisphere

Calculate y coordinates of the centroid as follows:

$$\begin{gathered} \overline{y}=\frac{\int yda}{\int da} \\ =\frac{{\int }_0^{\epsilon /2}{\int }_0^{2\pi }{a}^3\sin \succ d\succ {\sin }^2\theta d\theta }{{\int }_0^{*/2}{\int }_0^{2E}{a}^{2}d\succ \sin \theta d\theta } \\ =\frac{a^3{\int }_0^{2/2}\frac{(1-\cos 2\theta )}{2}d\theta -{\int }_0^{2x}\sin \succ d\succ }{{\left.a^2[-\cos \theta {]}_0^{2/2}\mid \succ \right]}_0^{2\alpha }} \\ =\frac{\frac{a}{2}{\left(\theta -\frac{\sin 2\theta }{2}\right)}_0^{\epsilon /2}·1-{\left.\cos \succ \right|}_0^{2x}}{-\left(\cos \frac{x}{2}-\cos 0\right)(2\pi -0)} \end{gathered}$$

Solve further, as follows:

$$\begin{gathered} \overline{y}=\frac{\frac{a}{2}\left(\frac{\pi }{2}-\sin \theta -0\right)(-1)(\cos 2\pi -\cos 0)}{2\pi (-1)(0-1)} \\ \overline{y}=0 \end{gathered}$$

Step 6: Calculation of   z coordinate of the centroid of the curved surface area of hemisphere

Calculate z coordinates of the centroid as follows:

$$\begin{gathered} \overline{z}=\frac{\int zda}{\int da} \\ =\frac{{\int }_0^{s/2}{\int }_0^{2\pi }{a}^{3}d\succ \sin \theta \cos \theta d\theta }{{\int }_0^{\epsilon /2}{\int }_0^{2x}{a}^{2}d\succ \sin \theta d\theta } \\ =\frac{a^3\frac{1}{2}{\int }_0^{x/2}\sin 2\theta d\theta ·{\int }_0^{2\pi }d\succ }{a^2{\int }_0^{x/2}\sin \theta d\theta } \\ =\frac{\frac{a}{2}{\left(\frac{\cos 2\theta }{2}\right)}_0^{\pi /2}}{[-\cos \theta {]}_0^{a/2}} \end{gathered}$$

Solve further, as follows:

$$\begin{gathered} \overline{z}=\frac{\frac{a}{4}(\cos \pi -\cos 0)}{\left(\cos \frac{\pi }{2}-\cos 0\right)} \\ =\frac{a}{4}\frac{(-1-1)}{0-1} \\ =\frac{a}{2} \end{gathered}$$

Thus, the centroid of curve surface area of hemisphere is $\left(0,0,\frac{a}{2}\right)$.