Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. `y = x sqrt(4-x^2)` , `y = 0` `` To calculate volume of solid generated by revolving a region bounded by `y=f(x)` and `x=a` and `x=b` we use the following formula

`V=pi int_a^b y^2 dx`

To find lower and upper bound we need to find intersections of the two curves i.e. we need to solve the following system of equations

`y=0`

`y=x sqrt(4-x^2)`

`x...

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To calculate volume of solid generated by revolving a region bounded by `y=f(x)` and `x=a` and `x=b` we use the following formula

`V=pi int_a^b y^2 dx`

To find lower and upper bound we need to find intersections of the two curves i.e. we need to solve the following system of equations

`y=0`

`y=x sqrt(4-x^2)`

`x sqrt(4-x^2)=0`

`x_1=0`

`4-x^2=0`

`x_2=-2`

`x_3=2`

So the curve intersects the x-axis at 3 points as you can see on the image below. So we can choose either lower part (from -2 to 0) or upper part (from 0 to 2). Since it is all the same I will choose the upper part (the other image). Therefore by using the above formula we get

`V=pi int_0^2(x sqrt(4-x^2))^2 dx=pi int_0^2(x^2(4-x^2))dx=`

`pi int_0^2(4x^2-x^2)dx=pi((4x^2)/3-x^5/5)|_0^2=pi(32/3-32/5)=(64pi)/15`

` `The volume is `(64pi)/15.`

The volume of rotating both upper and lower part of the curve would be twice as big i.e. `(128pi)/15.`

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