Determining the Moment of inertia

Let $a,b,c$ be positive real numbers such that $c<a$. Suppose given is a thin plate $R$ in the plane bounded by $$\frac{x}{a}+\frac{y}{b}=1, \frac{x}{c}+\frac{y}{b}=1, y=0$$ and such that the density of a point $(x,y) \in R$ is given by $\delta(x,y)=x$. Compute the mass of $R$

I found that the mass equals $\frac{(c^2-a^2)b}{6}$

However, I need to calculate as well the moments of inertia along the x-axsis and y-axsis.

I know that the moment of inertia along the x-axsis is given by:

$$I_x= \int \int_R y^2f(x,y)dxdy$$ and along the y-axis: $$I_y= \int \int_R x^2f(x,y)dxdy$$

But how do we use it here?

Thank you

• @DonAntonio Here it is – user43418 Apr 14 '13 at 18:21
• I'm not sure I understand the problem here: you say you already calculated the mass of the plate, and for that you had to compute an integral over $\,R\,$ , so what's new in this case? You already know the limits and stuff, don't you? So what is "it" in your last question?? – DonAntonio Apr 14 '13 at 18:24
• @DonAntonio The two above relation. How do I use what we did here: math.stackexchange.com/questions/361342/mass-of-a-rectangle/… – user43418 Apr 14 '13 at 18:24
• Maybe I'm missing something elementary or even obvious here, but aren't the above integrals exactly the same ones as in the link you presented but with $\,y^2\,,\,x^2\,$ resp.? – DonAntonio Apr 14 '13 at 18:32
• @DonAntonio So: $I_x=\int_0^b \int_{-\frac{c}{b}y+c}^{-\frac{a}{b}y+a} y^2dxdy$ and $I_y=\int_0^b \int_{-\frac{c}{b}y+c}^{-\frac{a}{b}y+a} x^2dxdy$ – user43418 Apr 14 '13 at 18:34

I'm going to refer to your last question taking, without checking, what you wrote there:

$$\int\limits_0^b\int\limits_{-\frac{c}{b}y+c}^{-\frac{a}{b}y+a}x^2dx\,dy=\frac{1}{3}\int\limits_0^b\left(-\frac{a^3}{b^3}(y-b)^3+\frac{c^3}{b^3}(y-c)^3\right)dy=$$

$$=\left.-\frac{a^3}{12b^3}(y-b)^4\right|_0^b+\left.\frac{c^3}{12b^3}(y-c)^4\right|_0^b=\ldots$$

I can't see any need to interchange the limits: as it is it's pretty easy to calculate it.

• No I was referring to the other one $I_x$ – user43418 Apr 14 '13 at 18:49
• It's even simpler, @user43418, as then you put the $\,y^2\,dy\,$ directly under the first integral and the second one is the integral of $\,1\cdot dx\,$ ...:P) – DonAntonio Apr 14 '13 at 18:53
• But it's still between 0 and b right ? I mean I integrate $y^2$ between 0 and b and then the other one right ? – user43418 Apr 14 '13 at 18:54
• @user43418 yes, no need to change anything else. – Coffee_Table Apr 14 '13 at 19:00

As in this other question, replace the mysterious $f$ with $\delta$ so that $$I_x= \int \int_R y^2\delta(x,y)dxdy,$$ and along the y-axis: $$I_y= \int \int_R x^2\delta(x,y)dxdy.$$ And again, $\delta$ has not changed from the other question, so simply substitute in its value into the above. Also the limits of integration have not changed from the other question! As you evaluate the inner integral with respect to $x$, you hold $y$ constant, so there's really no added difficulty from the other question. That is to say, in evaluating the inner part of $I_x$, you treat $y$ constant and so it poses no difficulty in the integration.