IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

avinash.sharma

iitkgp_bipin is right make it more clear by figures and description ( A general and perticular solution)

Figure 1 is the bounded area and figure 2 is the volume produced by rotating the area along x axis by 360 degree. Lets take a small cross section as in figure 3 of height h= dx having radius r=y then we can write the volume of such cylinder as

dV = p r²h = p y²dx

To calculate total volume we integrate this volume along dx when x moves from 0 to +1

V = ò p y²dx = ò p [f(x)]²dx where y =f(x) it is a general solution. In this integration you can put any y=f(x) and limits of x . In our case now put your curve y=x³ then

V = ò p [x³]²dx ( x moves 0 to +1 )

= ò p x⁶dx ( x moves 0 to +1 )

= p [x⁷/7] ( x moves 0 to +1 ) = p ( 1/7 ) = p/7

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