Question

Find the volume of the wedge with vertices at points (0,0,0), (1,0,0), (0,1,0), (0,0,1) by integrating the area of cross-sections parallel to the yz-plane. *

Answer 1

Answer:

V = 1/6 cubic units

Step-by-step explanation:

Applying the concept of integrals for volume calculation:

$V = \int\limits^b_a {S(x)} \, dx$          (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 = $S(x)=\frac{y(x)*z(x)}{2}$          (2)

Calculation of y(x) and z(x)

We apply the equation of the point-slope line (plane xy):

Slope = $m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}$          (3)

Equation of the line = $y - y_{1} =m(x-x_{1} )$          (4)

Replacing the points (1,0) and (0,1) in (3):

$m=\frac{1-0}{0-1} =-1$

Replacing the point (1,0) and m = -1 in (4):

$y-0=(-1)(x-1)$

y(x) = -x + 1 (Line A-B)          (5)

We apply the equation of the point-slope line (plane xz):

Slope = $m = \frac{z_{2} - z_{1} }{x_{2} - x_{1}}$          (6)

Equation of the line = $z - z_{1} =m(x-x_{1} )$          (7)

Replacing the points (1,0) and (0,1) in (6):

$m=\frac{1-0}{0-1} =-1$

Replacing the point (1,0) and m = -1 in (7):

$z-0=(-1)(x-1)$

z(x) = -x + 1 (Line A-C)        (8)

Replacing (5) and (8) in (2)

$S(x) = \frac{(-x + 1) * (-x + 1)}{2} =\frac{(-x + 1)^{2} }{2}$          (9)

Replacing (9) in (1) and knowing that a = 0 and b = 1:

$V = \int\limits^1_0 {\frac{(-x + 1)^{2} }{2}} \, dx = \int\limits^1_0 {\frac{x^{2}-2x+1 }{2}} \, dx$

$V =\frac{1}{2} (\frac{x^{3} }{3} -2\frac{x^{2} }{2} +x)$  evaluated from x=0 to x=1

$V= \frac{1}{2} (\frac{1}{3} -1 +1) = \frac{1}{6}$