Evaluating Moment of Area Integral using Geometry

[briteliner]

Homework Statement

the moment of area is integral (r^2 dA). With r measured from the origin, use geometry to evaluate this integral in both orders.

Homework Equations

The Attempt at a Solution

Ok so I set up the integral with dA=rdrd(theta) so with r from 0 to r and theta from 0 to pi/2. i ended up with pir^4/8 which i multiplied by for to account for each quadrant and ended up with (pir^4)/2.
is this correct? incorrect? why? i don't really get what the moment refers to. and also, if i am evaluating the correct integral. please help!

 

[tiny-tim]

Hi briteliner!

(have a pi: π and an integral: ∫ and try using the X² tag just above the Reply box )

the moment of area is integral (r^2 dA). With r measured from the origin, use geometry to evaluate this integral in both orders.

(this is the second moment of area … it measures a body's resistance to bending stress perpendicular to a particular axis )

What shape are you trying to find the moment of area of, and about what axis?

 

[briteliner]

a triangle, with x and y-axis i think

 

[tiny-tim]

a triangle, with x and y-axis i think

grrr! what shape of triangle?

Anyway, you don't need r and θ for a triangle …

just use dxdy as usual …

show us what you get. ​

 

[briteliner]

if b is the height and a is the width, i get (3(b^3)(x^3))/(y^3)

 

[tiny-tim]

if b is the height and a is the width, i get (3(b^3)(x^3))/(y^3)

(try using the X² tag just above the Reply box )

How can x and y be in the answer?

It should just have a and b.

(btw, is this a right-angled triangle? and you still haven't said what the axis is)

Anyway, show us your full calculations next time.

 

[flyfishing]

You're in Nearing's class, aren't you? lol

Pic is a right triangle with vertices at (0,0), (a,0), and (a,b).

Recall how to change the integrand so that you can integrate properly. (How is r related to x and y if you choose to integrate with respect to x & y?)