Find the moment of Inertia Bounded by the parabola $y^2 = 4x$, $x$-axis and $x=1$, with respect to the $x$-axis
The Answer is $1.067$
Formula for Moment of Inertia is:
$$Ix = \int_A y^2 dA$$
Finding Limits by Equating the Line and Parabola:
$$y^2 = 4(1)$$
$$y = \pm2$$
Integrate
$$Ix =\iint y^2 dxdy$$
$$Ix =\iint 4x dxdy$$
$$Ix =\int 2x^2 dy$$
$$Ix =\int 2\frac{y^4}{16} dy$$
$$Ix =\int_{-2}^2 2\frac{y^4}{16} dy$$
$$Ix = \frac{8}{5} = 1.6$$
Am I missing something? Any Hint?
Best Answer
$$ Ix = \iint_A y^2 dA = \int_{0}^{2}y^2\int^{1}_{y^2/4}dx\; dy =\int_{0}^{2} ( y^2 - \frac14 y^4 ) dy$$
$$ = \frac 13 y^3 - \frac{1}{20} y^5 \,\,|_{0}^{2} = \frac 83 - \frac{32}{20} $$
notice that the lower bound on the $y$ integration is $0$ because we are only looking at the part above the x-axis.
Your problem was that you were trying to use $y^2=4x$ in the integrand, but it actually only effects the limits of integration, when you are evaluating the integral you should treat the variables as if they were unrelated.