The area (in sq. units) of the region enclosed by the parabola y^2 =…

The correct answer is Option (1) → \(\displaystyle\frac{64}{3}\)

Writing both curves with x in terms of y, the parabola is \(x = \frac{y^{2}}{4}\) and the line is x = y + 3. The points of intersection come from

    \( \displaystyle \frac{y^{2}}{4} = y+3 \)

    \(y^{2}-4y-12 = 0\)
    \((y-6)(y+2) = 0\)

So y = −2 and y = 6, that is, the points (1, −2) and (9, 6). For y between −2 and 6 the line is to the right of the parabola (at y = 0, x = 3 on the line and x = 0 on the parabola). Integrating with respect to y avoids splitting the region:

    \( \displaystyle A = \int_{-2}^{6}\left(y+3-\frac{y^{2}}{4}\right)dy \)

    \( \displaystyle = \left[\frac{y^{2}}{2}+3y-\frac{y^{3}}{12}\right]_{-2}^{6} \)

At y = 6 and y = −2:

    \(18 + 18 - 18 = 18\)

    \( \displaystyle 2 - 6 + \frac{8}{12} = -\frac{10}{3} \)

So

    \( \displaystyle A = 18 + \frac{10}{3} = \frac{64}{3} \)