Volume of a tetrahedron

[MathsLearner]

I am trying to solve the below problem

My attempt is i am considering the other axis as z, not sure why it was not mentioned. Considering the right triangle 3,4, the area of the right triangle is
[MATH]\frac 1 2 yz[/MATH]. I need the equation of y and z in terms of x, and the x coordinates vary from 0 to 5. The equation of line1 is [MATH]y=-\frac 4 5 x + 4 [/MATH]The equation of line 2 is [MATH]z=-\frac 5 3 x + 3 [/MATH]. Hence the volume is [MATH] V = \int_0^5 \frac 1 2 (-\frac 4 5 x + 4)(-\frac 5 3 x + 3) [/MATH]. But the answer is not matching. What is the mistake? Please help.

 

[Deleted member 4993]

I am trying to solve the below problem
View attachment 18897
My attempt is i am considering the other axis as z, not sure why it was not mentioned. Considering the right triangle 3,4, the area of the right triangle is
[MATH]\frac 1 2 yz[/MATH]. I need the equation of y and z in terms of x, and the x coordinates vary from 0 to 5. The equation of line1 is [MATH]y=-\frac 4 5 x + 4 [/MATH]The equation of line 2 is [MATH]z=-\frac 5 3 x + 3 [/MATH]. Hence the volume is [MATH] V = \int_0^5 \frac 1 2 (-\frac 4 5 x + 4)(-\frac 5 3 x + 3) dx [/MATH]. But the answer is not matching. What is the mistake? Please help.

However, to catch your mistake, we need to see complete work including the answer you get and the answer given to you.

 

[MathsLearner]

The complete solution is
[MATH] \int_0^5(2-\frac {2x} 5)(3 - \frac{5x} 3)dx \\ \int_0^5(6 - \frac {10x} 3 - \frac {6x} 5 + \frac {10x^2} {15}) dx \\ \int_0^5(\frac {10x^2} {15} - \frac {68x} {15} + 6) dx \\ \int_0^5(\frac {2x^2} {3} - \frac {68x} {15} + 6) dx \\ [\frac {2x^3} 9 - \frac {34x^2} {15} + 6x]_0^5 \\ \frac {250} 9 - \frac {170} 3 + 30 =1.11 [/MATH]But the actual answer is 10.

 

[Deleted member 4993]

The complete solution is
[MATH] \int_0^5(2-\frac {2x} 5)(3 - \frac{5x} 3)dx \\ \int_0^5(6 - \frac {10x} 3 - \frac {6x} 5 + \frac {10x^2} {15}) dx \\ \int_0^5(\frac {10x^2} {15} - \frac {68x} {15} + 6) dx \\ \int_0^5(\frac {2x^2} {3} - \frac {68x} {15} + 6) dx \\ [\frac {2x^3} 9 - \frac {34x^2} {15} + 6x]_0^5 \\ \frac {250} 9 - \frac {170} 3 + 30 =1.11 [/MATH]But the actual answer is 10.

The equation of line 2 is z = 3 - 3/5 *x

Then everything will come together .... right now.....

 

[MarkFL]

I would consider (if using slicing) to use:

[MATH]dV=\frac{1}{2}bh\,dx[/MATH]
where

[MATH]b=-\frac{4}{5}x+4[/MATH]
[MATH]h=-\frac{3}{5}x+3[/MATH]
Hence:

[MATH]dV=\frac{1}{2}\left(-\frac{4}{5}x+4\right)\left(-\frac{3}{5}x+3\right)\,dx[/MATH]
[MATH]dV=\frac{6}{25}(5-x)^2\,dx[/MATH]
And so we have:

[MATH]V=\frac{6}{25}\int_0^5 (5-x)^2\,dx=\frac{6}{25}\int_0^5 u^2\,du=\frac{2}{25}\cdot5^3=10[/MATH]