Calculating the Area of a Half-Sphere Using Cylindrical Coordinates

Calculating the surface area of a parameterized surface like a half-sphere involves integrating the magnitude of the cross product of its partial derivative vectors, which is equivalent to the square root of the determinant of the induced metric tensor $( \left.g_{\alpha \beta}\right)$, where $g_{\alpha \beta}$ components are the dot products of the partial derivatives with respect to the parameters, and in this case, a cylindrical parametrization simplifies the integral by defining infinitesimal area elements ( $d S=\frac{R \rho}{\sqrt{R^2-\rho^2}} d \rho d \phi$ ) and allowing a u-substitution ( $u = R ^2- \rho ^2$ ) to yield the final result of $2 \pi R ^2$, confirming half the area of a full sphere.

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✍️Mathematical Proof

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Understanding Surface Area Integration

The fundamental concept is that the area of a parameterized surface is calculated by integrating the magnitude of the cross product of the partial derivative vectors of the parameterization. This magnitude is equivalent to the square root of the determinant of the induced metric tensor, $\sqrt{\operatorname{det}\left(g_{\alpha \beta}\right)}$.

The Induced Metric Tensor

The components of the metric tensor ( $g_{\alpha \beta}$ ) are found by taking the dot product of the partial derivatives of the position vector with respect to the parameters. In this problem, the position vector is $x(\rho, \phi)$, and the parameters are $\rho$ and $\phi$. So, $g_{\rho \rho}=\frac{\partial x}{\partial \rho} \cdot \frac{\partial x}{\partial \rho}, g_{\phi \phi}=\frac{\partial x}{\partial \phi} \cdot \frac{\partial x}{\partial \phi}$, and $g_{\rho \phi}=g_{\phi \rho}=\frac{\partial x}{\partial \rho} \cdot \frac{\partial x}{\partial \phi}$.

Calculation of the Area Element

The infinitesimal area element $d S$ is given by $d S= \sqrt{\operatorname{det}\left(g_{\alpha \beta}\right)} d \rho d \phi$. In this specific problem, the result is $d S=\frac{R \rho}{\sqrt{R^2-\rho^2}} d \rho d \phi$.

Integration with a u-substitution

The integral over $\rho$ requires a u-substitution, which is a common technique for solving integrals involving square roots of quadratic expressions. The substitution $u=R^2-\rho^2$ simplifies the integral significantly.

Final Result Consistency

The final result, $2 \pi R^2$, correctly corresponds to half the surface area of a full sphere, which is $4 \pi R^2$. This consistency check provides confidence in the accuracy of the calculation.

✍️Mathematical Proof

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🧄Proof and Derivation-2

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