Question:
The volume of the tetrahedron bounded by the surfaces \(x = 0\), \(y = 0\), \(z = 0\) and \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\) is:
सतहों \(x = 0\), \(y = 0\), \(z = 0\) और \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\) द्वारा परिबद्ध चतुष्फलक (tetrahedron) का आयतन क्या है?
Options:
- \(\frac{abc}{4}\)
- \(\frac{abc}{2}\)
- \(\frac{abc}{10}\)
- \(\frac{abc}{6}\)
- \(\frac{abc}{4}\)
- \(\frac{abc}{2}\)
- \(\frac{abc}{10}\)
- \(\frac{abc}{6}\)
Correct Answer: d)
Explanation:
The volume \(V\) of a tetrahedron bounded by the three coordinate planes and the plane \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\) is evaluated via the triple integral:
\[ V = \int_{0}^{a} \int_{0}^{b(1 – x/a)} \int_{0}^{c(1 – x/a – y/b)} dz \, dy \, dx = \frac{1}{6} abc \]
Geometrically, this represents a pyramid with base area in the \(xy\)-plane of \(\frac{1}{2}ab\) and height \(c\):
\[ V = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times \left(\frac{1}{2}ab\right) \times c = \frac{abc}{6} \]
निर्देशांक तलों \(x=0, y=0, z=0\) और समतल \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\) द्वारा परिबद्ध चतुष्फलक का आयतन:
\[ V = \frac{1}{3} \times \text{आधार का क्षेत्रफल} \times \text{ऊँचाई} = \frac{1}{3} \times \left(\frac{1}{2}ab\right) \times c = \frac{abc}{6} \]