Question

Find the centroid of the solid generated if the region bounded by y = 2- x^2, x=0, and y=0 is revolved about the y-axis.

Answer 1

Solution:

Given, $f(x)=y=2-x^2$

Since, it is revolving around y-axis, centroid will lie on y-axis.

So, it is of the form $C(0,y,0)$.

$y=\dfrac{\int_0^bf^2(x)dx}{2\int_a^bf(x)dx} \\=\dfrac{\int_0^{\sqrt{2}}(2-x^2)^2dx}{2\int_0^{\sqrt{2}}(2-x^2)dx}$

$=\dfrac{\int_0^{\sqrt{2}}(4+x^4-4x^2)dx}{2\int_0^{\sqrt{2}}(2-x^2)dx}$

$=\dfrac{(4x+x^5/5-4x^3/3)_0^{\sqrt{2}}}{2(2x-x^3/3)_0^{\sqrt{2}}}$

$=\dfrac{(4{\sqrt{2}}+({\sqrt{2}})^5/5-4({\sqrt{2}})^3/3)}{2(2{\sqrt{2}}-({\sqrt{2}})^3/3)}$

$=\dfrac{(4{\sqrt{2}}+(4{\sqrt{2}})/5-8({\sqrt{2}})/3)}{2(2{\sqrt{2}}-(2{\sqrt{2}})/3)}$

$=\frac45$

Now, $C(0,\frac45,0)$