# What is The volume of the solid generated when the region bounded by the curve y = sqrt(x) and y = x revolved about the line x = -2

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### 1 Answer

We want to find the volume of the solid generated when the region between the graphs of `y=sqrt(x)` and y=x is rotated about the line x=-2:

Using the shell method we have `V=2piint_a^bp(x)h(x)dx`

Here a=0,b=1 (the graphs intersect at x=0,x=1), p(x) is the distance of the center of the rectangle from the vertical axis of rotation which is x+2, and h(x) is the height of the rectangle at x which is `h(x)=sqrt(x)-x` .(The graph of square root of x is above the line y=x on the interval (0,1).)

Then `V=2piint_0^1(x+2)(sqrt(x)-x)dx`

`=2pi[int_0^1 x^(3/2)dx-int_0^1x^2dx +2int_0^1x^(1/2)-2int_0^1xdx]`

`=2pi[2/5x^(5/2)|_0^1-x^3/3|_0^1+4/3x^(3/2)|_0^1-x^2|_0^1 ]`

`=2pi(6/15-5/15+20/15-15/15)`

`=(4pi)/5`

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**The volume is `(4pi)/5` cubic units.**

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