Sketch the region lying in the first quadrant and bounded by `y=9x^2, x=0,y=1a n dy=4.` Find the area of the region using integration.

Explanation

Here, `x^{2}=frac{y}{9}` is a upward parabola having its vertex at the origin and `y=1, y=4` are the lines which are parallel to `{x}`-axis at `{y}=1` to `{y}=4` unit distance from it
Similarly `x^{2}=frac{y}{9}` has even power of `y` and is symmetrical about `y-` axis
So the required area is written as


Required area `=int_{1}^{4} {xdy}`



`=int_{1}^{4} frac{sqrt{y}}{3} d y`




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