# Calculate surface area of a sphere using the surface integral

Given a sphere with: $$F := \{(x,y,z) \in \mathbb{R}^3 \mid x^2+y^2+z^2 = 1, x\le0\}$$ $$\Rightarrow r = 1, \varphi = [\frac{\pi}{2}, \frac{3\pi}{2}], \theta = [0, \pi]$$

My Task is to calculate the surface area by using surface integral.

The default sphere paramterization: $$X\left(\varphi,\theta\right)= \begin{pmatrix} \cos \varphi\sin \theta\\ \sin \varphi\sin \theta\\ \cos \theta \end{pmatrix}$$

And it's derivates: $$X_{\varphi} \times X_{\theta} = \begin{pmatrix} -\sin \varphi\sin\theta\\ \cos \varphi\sin\theta\\ 0 \end{pmatrix} \ \ \times \ \ \begin{pmatrix} \cos \varphi\cos\theta\\ \sin \varphi\cos\theta\\ -\sin\theta \end{pmatrix}\ = \begin{pmatrix} -\cos\varphi\sin^2\theta\\ -\sin\varphi\sin^2\theta\\ -\sin\theta\cos\theta \end{pmatrix}$$

The euclidean norm: $$\| \begin{pmatrix} -\cos\varphi\sin^2\theta\\ -\sin\varphi\sin^2\theta\\ -\sin\theta\cos\theta \end{pmatrix} \| = \sin^2\theta$$

$$S(F)=\int_0^{\pi}\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\sin^2\theta \; d\varphi d\theta = \frac{\pi^2}{2}$$

Now my question is, where did I make a mistake? Since the surface should the surface of half a sphere, being $1/2 * 4\pi * 1^2 = 2\pi$

• Shouldn't write $X(u,y)$ when you do not use $u,v$ (that's not an answer). – GDumphart Dec 18 '14 at 12:38
• I get the norm to be $\lvert \sin \theta\rvert$ and not $\sin^2\theta$. – ronno Dec 18 '14 at 12:58
• @ronno yeah you are right, and since $\theta$'s range is $[0, \pi]$ we can even omit the absolute. – L1me Dec 18 '14 at 13:13

$$\left\| \begin{pmatrix} -\cos\varphi\sin^2\theta\\ -\sin\varphi\sin^2\theta\\ -\sin\theta\cos\theta \end{pmatrix} \right\| = \sqrt{(\cos^2\varphi+\sin^2\varphi)\sin^4\theta + \sin^2\theta\cos^2\theta} = \sqrt{(\sin^2\theta + \cos^2\theta)\sin^2\theta} = \sqrt{\sin^2\theta} = \sin\theta$$
$$S(F)= \left(\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}d\varphi\right) \cdot \left( \int_0^{\pi} \sin\theta d\theta\right) = \pi \cdot 2 = 2\pi$$