Question

Find the volume of the solid formed by revolving the region bounded by y=(x-2)² and y=x about the y -axis.

Answer 1

Find the points of intersection of the parabola with the y=x

$x = (x-2)^2$

We get $x = 1,4$

As the region is revolved about the y axis, we express the equation of the bounding curve in terms of y

$y = (x-2)^2$

$(x-2) =\pm \sqrt{y}$

$x = 2 \pm \sqrt{y}$

Volume can be calculated as:

$V = \pi \int_{1}^{4}[(2+\sqrt{y})^2-(2-\sqrt{y})^2]dy$

$V = \pi \int_{1}^{4}4\sqrt{y}dy$

$V = \dfrac{8\pi}{3}[y^{3/2}]_{1}^{4}$

$V = \dfrac{56\pi}{3}$