Questions: 7.1) จงหาพื้นที่ผิวซึ่งเป็นส่วนหนึ่งของทรงกระบอกเชิงกลม (x^2+z^2=9) ในอัฐภาค (octant) ที่ 1 ชึ่งถูกตัดด้วยระนาบ (x=0), (x=3), (y=0) และ (y=5)

Solution Steps

To find the surface area of a portion of a cylindrical surface in the first octant, we need to consider the given boundaries. The cylinder is defined by \(x^2 + z^2 = 9\), and we are interested in the part of this cylinder that lies between the planes \(x=0\), \(x=3\), \(y=0\), and \(y=5\). We will set up an integral to calculate the surface area over this region.

The cylinder is defined by the equation \( x^2 + z^2 = 9 \). We are interested in the portion of this cylinder that lies in the first octant, bounded by the planes \( x = 0 \), \( x = 3 \), \( y = 0 \), and \( y = 5 \).

From the cylinder equation, we can express \( z \) in terms of \( x \): \[ z = \pm \sqrt{9 - x^2} \] In the first octant, we take the positive root, so \( z = \sqrt{9 - x^2} \).

The surface area element \( dS \) can be calculated using the formula: \[ dS = \sqrt{1 + \left( \frac{dz}{dx} \right)^2} \, dy \, dx \] We find \( \frac{dz}{dx} \): \[ \frac{dz}{dx} = \frac{x}{\sqrt{9 - x^2}} \] Thus, the surface area element becomes: \[ dS = \sqrt{1 + \frac{x^2}{9 - x^2}} \, dy \, dx = \sqrt{\frac{9}{9 - x^2}} \, dy \, dx \]

We need to integrate \( dS \) over the specified limits for \( x \) and \( y \): \[ \text{Surface Area} = \int_{0}^{3} \int_{0}^{5} \sqrt{\frac{9}{9 - x^2}} \, dy \, dx \] Calculating the inner integral with respect to \( y \) gives: \[ \int_{0}^{5} dy = 5 \] Thus, the surface area simplifies to: \[ \text{Surface Area} = 5 \int_{0}^{3} \sqrt{\frac{9}{9 - x^2}} \, dx \]

The integral can be evaluated, yielding: \[ \text{Surface Area} = 15 \cdot \frac{\pi}{2} = \frac{15\pi}{2} \]

Final Answer