Calculus problem #2046

Calculus Level 4

A thin triangle has vertices ( 0 , 0 ) (0, 0) , ( 1 , 0 ) (1, 0) and ( 0 , 2 ) . (0, 2). Let the density function at any given point on the triangle be represented by ρ ( x , y ) = 1 + 3 x + y \rho (x, y) = 1+3x+y . What is the mass of this triangle?

Details and assumptions

The density of a material is defined as ρ = m V \rho = \frac{m}{V} , where ρ \rho is the density, m m is the mass and V V is the volume.


The answer is 2.6666666.

Calvin Lin by Calvin Lin

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2 solutions

Kenny Lau
Nov 29, 2014

m = ∫ 0 1 ∫ 0 2 − 2 x ( 1 + 3 x + y ) d y d x = ∫ 0 1 [ y + 3 x y + y 2 2 ] 0 2 − 2 x d x = ∫ 0 1 ( 2 − 2 x ) + 3 x ( 2 − 2 x ) + ( 2 − 2 x ) 2 2 d x = ∫ 0 1 2 − 2 x + 6 x − 6 x 2 + 4 x 2 − 8 x + 4 2 d x = ∫ 0 1 2 + 4 x − 6 x 2 + 2 x 2 − 4 x + 2 d x = ∫ 0 1 − 4 x 4 + 4 d x = [ − 4 3 x 3 + 4 x ] 0 1 = 8 3 \begin{array}{rcl} m&=&\int_0^1\int_0^{2-2x}(1+3x+y)dydx\\ &=&\int_0^1\left[y+3xy+\frac{y^2}2\right]_0^{2-2x}dx\\ &=&\int_0^1(2-2x)+3x(2-2x)+\frac{(2-2x)^2}2dx\\ &=&\int_0^12-2x+6x-6x^2+\frac{4x^2-8x+4}2dx\\ &=&\int_0^12+4x-6x^2+2x^2-4x+2dx\\ &=&\int_0^1-4x^4+4dx\\ &=&\left[-\frac43x^3+4x\right]_0^1\\ &=&\frac83\\ \end{array}

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Sundar R
Feb 9, 2014

The equation of the hypotenuse line is y=-2x+2

Now we integrate the density function y=(1+3x+y) as a function of y holding x constant (say xo) with integration limits going from y=0 to y=2-2xo

Now we have an equation in terms of xo which we integrate from 0 to 1 (the segment of the length along the x axis)

Finally , we multiply by the area to get the total mass

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