We recognise that this is an ellipse. The question tells us the area of interest is in the first quadrant only.
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[4-4/3] - pi`