Respuesta :
Answer:
8 unit^3
Step-by-step explanation:
Given:
- The equation of the plane is:
3x + 6y + 2z = 12
Find:
Use a double integral to find the volume of the solid in the first octant which is enclosed by the plane and the coordinate planes
Solution:
- Express the equation of surface ( plane ) as a subject of any coordinate axis we will use z:
2z = 12 - 3x - 6y
z = 6 - 1.5x - 3y
- The double integral would be set- up as:
[tex]\int\limits^d_c \int\limits^a_b ({6 - 1.5x - 3y}) \, dy.dx[/tex]
- Where, a , b ,c and d are limits of integration.
- To determine the limits we will project the surface to x-y plane or z = 0 plane, the equation we have is:
0 = 6 - 1.5x - 3y
y = 2 - 0.5x
- For limits a and b the integration is with respect to y, so we express the limits of y in terms of x. Where lower limit b = 0, and upper limit a = 2 - 0.5x
- Similarly, the limits c and d is with respect to x are constants we have:
c = 0
0 = 2 - 0.5*d
d = 4
- Then solve the double integral:
[tex]\int\limits^4_0 ({6y - 1.5xy - 1.5y^2}) \,_0 ^2^-^0^.^5^x dx \\\\\int\limits^4_0 ({6(2-0.5x) - 3x + 0.75x^2 - 1.5(2-0.5x)^2}) dx \\\\({-6(2-0.5x)^2 - 1.5x^2 +0.25x^3 + (2-0.5x)^3}) | ^4_0\\\\= ( -6(0) - 1.5(16) + 0.25*(64) + (0) + 6(4) + 0 + 0 - (8) ) \\\\= 8 unit^3[/tex]
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