Answer:
V = 1/6 cubic units
Step-by-step explanation:
Applying the concept of integrals for volume calculation:
[tex]V = \int\limits^b_a {S(x)} \, dx[/tex] (1)
V = volume of the solid bounded by x = a and x = b
S(x) = cross section area of the solid, perpendicular to the x axis
From the figure we have that S is the area of a triangle that has base Z and height Y
Area of the triangle = [tex]S(x)=\frac{y(x)*z(x)}{2}[/tex] (2)
Calculation of y(x) and z(x)
We apply the equation of the point-slope line (plane xy):
Slope = [tex]m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}[/tex] (3)
Equation of the line = [tex]y - y_{1} =m(x-x_{1} )[/tex] (4)
Replacing the points (1,0) and (0,1) in (3):
[tex]m=\frac{1-0}{0-1} =-1[/tex]
Replacing the point (1,0) and m = -1 in (4):
[tex]y-0=(-1)(x-1)[/tex]
y(x) = -x + 1 (Line A-B) (5)
We apply the equation of the point-slope line (plane xz):
Slope = [tex]m = \frac{z_{2} - z_{1} }{x_{2} - x_{1}}[/tex] (6)
Equation of the line = [tex]z - z_{1} =m(x-x_{1} )[/tex] (7)
Replacing the points (1,0) and (0,1) in (6):
[tex]m=\frac{1-0}{0-1} =-1[/tex]
Replacing the point (1,0) and m = -1 in (7):
[tex]z-0=(-1)(x-1)[/tex]
z(x) = -x + 1 (Line A-C) (8)
Replacing (5) and (8) in (2)
[tex]S(x) = \frac{(-x + 1) * (-x + 1)}{2} =\frac{(-x + 1)^{2} }{2}[/tex] (9)
Replacing (9) in (1) and knowing that a = 0 and b = 1:
[tex]V = \int\limits^1_0 {\frac{(-x + 1)^{2} }{2}} \, dx = \int\limits^1_0 {\frac{x^{2}-2x+1 }{2}} \, dx[/tex]
[tex]V =\frac{1}{2} (\frac{x^{3} }{3} -2\frac{x^{2} }{2} +x)[/tex] evaluated from x=0 to x=1
[tex]V= \frac{1}{2} (\frac{1}{3} -1 +1) = \frac{1}{6}[/tex]
