One more question! CALC 2

[andoverhockey]

Find the total mass of the triangular region shown below. All lengths are in centimeters, and the density of the region is [FONT=MathJax_Math-italic]δ[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]1[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]x[/FONT][/FONT]grams/cm[FONT=MathJax_Main]2[/FONT].



[FONT=Helvetica Neue, Helvetica, Arial, sans-serif]I know:[/FONT]

[FONT=Helvetica Neue, Helvetica, Arial, sans-serif]Mass = density x area[/FONT]

[FONT=Helvetica Neue, Helvetica, Arial, sans-serif]I know my teacher said to take two separate integrals like:[/FONT]
[FONT=Helvetica Neue, Helvetica, Arial, sans-serif]1)[/FONT]
[FONT=Helvetica Neue, Helvetica, Arial, sans-serif]a = -1[/FONT]
[FONT=Helvetica Neue, Helvetica, Arial, sans-serif]b = 0

[/FONT]the integral from -1 to 0 of (1+x)... then what do I do with the area?

[FONT=Helvetica Neue, Helvetica, Arial, sans-serif]2)[/FONT]
[FONT=Helvetica Neue, Helvetica, Arial, sans-serif]a = 0[/FONT]
[FONT=Helvetica Neue, Helvetica, Arial, sans-serif]b = 1

[/FONT]the integrand from 0 to 1 of (1+x)...

I bet I am over thinking thinking this.

 

[Ishuda]

Just a quick statement - you need the equations of the lines forming the sides of the triangle and the equation is not y = x + 1.

 

[HallsofIvy]

Since the density depends only on "x", do it by thinking of thin vertical rectangles. As Ishuda pointed out, the equation of the top boundary is different on left and right of the y-axis so you need to do this in two parts, to the left of the y-axis and to the right.

On the left, the line goes from (-1, 0) to (0, 3) so has slope (3- 0)/(0-(-1))= 3 so has equation y= 3(x+ 1). Each "thin vertical rectangle", centered at x with width dx, has area 3(x+ 1)dx= (3x+ 3)dx. Its mass is "area times density" so is (1+ x)(3x+ 3)dx= (3x^2+ 6x+ 3)dx. Integrate that from x= -1 to x= 0.

On the right, the top boundary is a line from (0, 3) to (1, 0) so has slope (0- 3)/(1- 0)= -3 and so equation y= -3(x- 1)= -3x+ 3. The mass of a "thin vertical rectangle", centered at x with width dx, is (1+ x)(-3x+ 3)dx= (-3x^2+ 3)dx. Integrate that from x= 0 to x= 1.