We recognise that this is an ellipse. The question tells us the area of interest is in the first quadrant only.

121xyyΔxy = f(x)Open image in a new page

The ellipse x2 + 4y2 = 4, showing the portion bounded by the curve, x = 0, x = 2 and the x-axis.

From the diagram, we can see that the limits of the bounded area are y = 0 and y = 1.

Hence the volume generated is:

`"Vol"=pi int_c^d x^2 dy`

`=pi int_0^1 (4-4y^2) dy`

`=pi[4y-(4y^3)/(3)]_0^1`

`=pi[4-4/3] - pi[0]`

`=(8pi)/(3)\ text[units]^3`

`~~8.378\ text[units]^3`

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