# If the curve x^2 + y^2 - 4 = 0 is rotated by 360 degrees about the x-axis what is the volume of the solid created.

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Solve the equation for y

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

`A=int^2_-2 pi(sqrt(4-x^2))^2 dx = int^2_-2 pi(4-x^2)dx = pi(4x-1/3x^2)|^2_-2`

`A=pi(4(2)-2^3/3-4(-2)+(-2)^3/3)=pi(16-2/3(8))=pi(16-16/3)`

`A=pi(32/3)=(32pi)/3`

The curve x^2 + y^2 - 4 = 0 is rotated about the x-axis. It is assumed you are referring to the volume enclosed by the shell formed by the curve as it is rotated 360 degrees about the x-axis.

x^2 + y^2 - 4 = 0

=> x^2 + y^2 = 2^2

This is the equation of a circle with center (0, 0) and radius 4. The x-axis or y = 0 is a line along a diameter of the circle. If the circle is rotated about this line by 360 degrees the resulting shape is that of a sphere with radius equal to 2.

The volume of a sphere with radius 2 is given by `V = (4/3)*pi*r^3`

Substituting r = 2

`V = (4/3)*pi*2^3`

=> `V = (4/3)*pi*8`

=> `V = (32/3)*pi`

**The volume of the solid enclosed by the curve x^2 + y^2 - 4 = 0 where it is rotated about the x-axis is `(32/3)*pi` **

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