Use a triple integral to find the volume of the given solid. the solid enclosed by the cylinder the result is: 8/15
Triple integrals are the three-dimensional equivalent of double integrals.
They are used to sum up an unlimited number of minuscule quantities connected with locations in a three-dimensional area.
solving planes z=0 and y+z=1
we have y=1
now solving plane y=1 and the given cylinder y=x^2
we have x^2=1
so x = +/- 1
so limits of Z are ( 0 to 1-y)
limits of y are (x^2 to 1)
limits of x are (-1 to 1)
so VOLUME = [tex]\(\int_{-1}^{1}\int_{x^2}^{1}\int_{0}^{1-y} dzdydx\)[/tex]
on integrating wrt dz we have z and between limits (1-y ,0) we have (1-y)
so we have [tex]\(\int_{-1}^{1}\int_{x^2}^{1}(1-y)dydx\)[/tex]
so on integration wrt to dy we have -1*((1-y)^2/2) and between the limits x^2 and 1
we have [tex]-1*((1-x^2)/2 - 0) = (-1/2)*(1-x^2)^2[/tex]
so we now have [tex]\(\int_{-1}^{1}(-1/2)*(1-x^2)^2 dx\)[/tex]
on integrating we have 8/15.
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