Respuesta :
Answer:
2.5 units^3
Step-by-step explanation:
Given:-
- The solid is bounded by a plane defined by the following points:
P(3, 0, 0), Q(0, 1, 0),R(0, 0, 2)
Find:-
Use a double integral to find the volume of the solid in the first octant bounded by the plane
Solution:-
- Determine the equation of the plane. Compute two direction vectors d1 and d2 that lie on the plane:
d1 = P - Q
d1 = (3, 0, 0) - (0, 1, 0) = (3,-1,0)
d2 = P - R
d2 = (3, 0, 0) - (0, 0, 2) = (3,0,-2)
- Find the a vector "normal" - n to the plane by cross product formulation of direction vectors (d1 and d2) that lie on the plane:
[tex]n = d1xd2 = \left[\begin{array}{c}3&-1&0\end{array}\right] x \left[\begin{array}{c}3&0&-2\end{array}\right] = \left[\begin{array}{ccc}3&-1&0\\3&0&2\end{array}\right] = \left[\begin{array}{c}-2&-6&3\end{array}\right][/tex]
- The equation of plane is:
n.(x,y,z) = n.P
-2x -6y + 3z = -6
- The function of one variable would be:
z = (2/3)x + 2y - 2
- The double integration formulation would be:
[tex]\int\limits^a_b \int\limits^c_d f(z) dy.dx\\\\\int\limits^a_b \int\limits^c_d (\frac{2x}{3} + 2y - 3) dy.dx\\\\\int\limits^a_b (\frac{2xy}{3} + y^2 - 3y) |_c^d.dx[/tex]
- Where the limits (c and d) are defined by planar (x-y) projection of plane (n) :
y = d = -(1/3)x + 1
c = 0
- Evaluate the limits:
[tex]\int\limits^a_b (\frac{-2x^2 + 6x}{9} + \frac{x^2}{9} -\frac{2x}{3} +1 + x - 3) .dx\\\\\int\limits^a_b (\frac{-x^2 }{9} + x - 2) .dx\\\\(\frac{-x^3 }{27} + \frac{x^2}{2} - 2x)|^3_0\\\\(\frac{-27 }{27} + \frac{9}{2} - 6) = 1 - 4.5 + 6 = 2.5 unit^3[/tex]
