Use a triple integral to find the volume of the given solid. The solid enclosed by the cylinder y = x^2 and the planes z = 0 and y + z = 1.
y = x2
z = 0 and y + z = 1
The limits of x, y and z are
x = -1 to 1
y = x2 to 1
z = 0 to 1 - y
Let x2 = k
The volume of the cylinder is given by
V = "\\int"-11 "\\int" k 1 "\\int" 01-y dz dy dx
V = "\\int"-11 "\\int" k1 ( 1 - y ) dy dx
V = "\\int"-11 ( y - "\\dfrac{y^2}{2}" )k 1 dx
V = "\\int"-11 ( 1 - k - "\\dfrac{1}{2}" + "\\dfrac{k^2}{2}" ) dx
On substituting the value of k, we have
V = "\\int"-11 ( 1 - x2 - "\\dfrac{1}{2}" + "\\dfrac{x^4}{2}" ) dx
V = "\\int"-11 ( "\\dfrac{1}{2}" - x2 + "\\dfrac{x^4}{2}" ) dx
V = ( "\\dfrac{1}{2}"x - "\\dfrac{x^3}{3}" + "\\dfrac{x^5}{10}" )-1 1
V = [ "\\dfrac{1}{2}"(1 + 1) - "\\dfrac{(1+1)^3}{3}" + "\\dfrac{(1+1)^5}{10}" ]
V = [ 1 - "\\dfrac{8}{3}" + "\\dfrac{32}{10}" ]
V = 1.533 cubic units.
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