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` ``

1 Answer | Add Yours

tiburtius's profile pic

tiburtius | High School Teacher | (Level 2) Educator

Posted on

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.` 

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 2)
This image has been Flagged as inappropriate Click to unflag
Image (2 of 2)

We’ve answered 318,992 questions. We can answer yours, too.

Ask a question