Calc 3 Moments and centers of mass

[Litcyb]

Homework Statement

Find the center of mass of a lamina in the shape of an isosceles right triangle with equal sides of length a if the density at any point is proportional to the square of the distance from the vertex opposite the hypotenuse.

Homework Equations

∫∫ (f(x,y) dA
mx= 1/m(∫∫ x(fx,y) dA
my= 1/m(∫∫ y(f(x,y) dA

The Attempt at a Solution

how do you calculate the bounds?

I know its the distance from the 90° angle to the the hypotenuse, but how to calculate that length? According to the book that length is √(x^2+y^2) why? Please, Help me how to visualize this problem.

I know how to calculate the density and centers of mass, I am just struggling in visualizing the problem and coming with an equation to integrate. Thank you. Ps- Happy Thanksgiving to those who celebrate it.

 

[Litcyb]

omg, you just use the distance formula! from (0,0) (90°) to any point on the line C (hypotenuse (x,y). How could i not remember that...not bueno.

I was able to calculate the bounds also! So happy, the bounds are from 0 to a, for the outer integral and 0 to a-x for the inner integral! we just draw the picture and cut it in perspective of x, so dA will now be, dydx if you cut it horizontally, parallel to the x-axis.

 

[HallsofIvy]

I presume you have taken the right angle at the origin and the hypotenuse in the first quadrant. In that case, yes, the distance from any point (x, y) to the right angle, the origin, is [itex]\sqrt{x^2+ y^2}[/itex], from the usual distance formula.

Yes, taking the length of the other two sides as "a", the other two vertices are at (a, 0) and (0, a) and the equation of the hypotenuse is given x+ y= a or y= a- x. The "outer" integral must be from 0 to a and the "inner" integral from 0 to a- x. Taking the "proportion" to be k, the mass of this lamina is
[tex]\int_{x=0}^a\int_{y=0}^{a-x} k\sqrt{x^2+ y^2}dydx[/tex]
Is that what you got?