2. Set up but do not evaluate the integral representing the volume of the region (circular ellipsoid) bounded by the ellipse ax^2+by^2=r^2 about the x-axis.

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You need to determine the area of the washer, hence, you need to evaluate the inner and outer radii, such that:

`ax^2 + by^2 = r^2 => by^2 = r^2 - ax^2 => y^2 = (r^2 - ax^2)/b`

`y_(1,2) = +-sqrt((r^2 - ax^2)/b)`

Evaluating the area of the washer yields:

`A = pi*(sqrt((r^2 - ax^2)/b))^2 - 0`

`A = pi*(r^2 - ax^2)/b^2`

Setting up the volume, yields:

`V = int_(-r)^r pi*(r^2 - ax^2)/b^2 dx=> V = (pi/b^2)*int_(-r)^r (r^2 - ax^2) dx`

Hence, setting up the integral that helps you to evaluate the volume of the requested region, yields `V = (pi/b^2)*int_(-r)^r (r^2 - ax^2) dx` .

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