Posted by **Anonymous** on Monday, February 13, 2012 at 10:18am.

Find the surface area of the solid generated by rotating the area between the y-axis, (x^2/y) + y = 1, and -1≤y≤0 is rotated around the y-axis.

- calculus -
**Anonymous**, Monday, February 13, 2012 at 11:10am
you have a circle:

x^2 + (y-1/2)^2 = 1/4

except that as originally written, y cannot be zero.

Now, since the range of y is (0,1], there is no part of the graph below the x-axis. Note that any y < 0 will produce a value such that (y-1/2)^2 > 1/4. That means that x^2 must be negative.

Is this a trick question?

If there's a typo, and you want 0<y<=1, then you are just rotating a semi-circle of radius 1/2, whioch will give you a sphere.

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