The equation of the line that passes throughout the points (3,0,0) and (0,4,0) in the plane xOy is 3x+4y=1 and hence y=4(1−3x). The equation of the plane that passes throughout the points (3,0,0), (0,4,0) and (0,0,5) is 3x+4y+5z=1 and thus z=5(1−3x−4y). It follows that the volume V is uqual to
V=∫03dx∫04(1−3x)dy∫05(1−3x−4y)dz=∫03dx∫04(1−3x)5(1−3x−4y)dy=5∫03(4(1−3x)−43x(1−3x)−81(4(1−3x))2)dx=5∫03(4−34x−43x+94x2−2+43x+29x2)dx=5∫03(2−34x+23x2)dx=5(2x−32x2+92x3)∣03=5(6−6+2)=10
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