Given y=sqrt(x) ,and to find the volume of the solid of rotation about x-axis,So, this can be solved using the shell method. And is as follows,y=sqrt(x) =>x=y^2 and the 0<=y<=1 these are the limits of ynow in the shell method the volume is given as ,V= 2*pi int _c ^d p(y)h(y) dy here p(y) is the average radius about x axis =yand h(y) is the function of height=y^2 and c=0 and d=1 as the range of yso, the volume is:V=2*pi int _c ^d p(y)h(y) dy =2*pi int _0 ^1 (y)(y^2) dy =2*pi int _0 ^1 (y^3) dy =2*pi [y^4/4] _0 ^1 =2*pi[[1/4]-[0]] =pi/2 is the volume
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