Question 12.6: A right triangle with base b and height h is shown in Fig. 1......

(a) Product of inertia with respect to the xy axes. We will use the method of integration to evaluate this product of inertia. We begin by considering a differential element of area dA (Fig. 12-23) in the form of a thin, horizontal strip of height dy and width equal to

\frac{(h\ -\ y)b}{h}

The area of this elemental strip is

dA=\frac{(h\ -\ y)b}{h}dy

and the coordinates of its centroid with respect to the xy axes are

d_1=y \quad \quad d_2=\left(\frac{1}{2}\right)\frac{(h\ -\ y)b}{h}

in which d_1 and d_2 are defined in Fig. 12-21. The product of inertia of the strip with respect to axes through its own centroid and parallel to the xy axes is zero (from symmetry). Therefore, its product of inertia dI_{xy} with respect to the xy axes (from the parallel-axis theorem, Eq. 12-20) is

dI_{xy}=0+(dA)d_1d_2=\left[\frac{(h\ -\ y)b}{h}dy\right]\left[y\left(\frac{1}{2}\right)\frac{(h\ -\ y)b}{h}\right]=\frac{b^2}{2h^2}(h\ -\ y)^2y\ dy

The product of inertia I_{xy} of the entire triangle is obtained by integration:

I_{xy}=\int{dI_{xy}}=\frac{b^2}{2h^2}\int_{0}^{h}{(h\ -\ y)^2\ dy}=\frac{b^2h^2}{24}                         (12-22)

Note that the entire area lies in the first quadrant, and therefore the product of inertia is positive.
(b) Product of inertia with respect to the x_cy_c axes. The product of inertia with respect to axes through the centroid may be determined from the parallel-axis theorem (Eq. 12-20):

I_{x_cy_c}=I_{xy}\ -\ A\left(\frac{h}{3}\right)\left(\frac{b}{3}\right)=\frac{b^2h^2}{24}\ -\ \frac{bh}{2}\left(\frac{h}{3}\right)\left(\frac{b}{3}\right)=-\frac{b^2h^2}{72}             (12-23)

in which h/3 and b/3 are the coordinates of point C with respect to the xy axes. Since most of the area is located in the second and fourth quadrants, the product of inertia turns out to be negative. The products of inertia given by Eqs. (12-22) and (12-23) are listed in Cases 7 and 6, respectively, of Appendix D.