Hi, I am new to this forum. Greetings to everyone. I'm kind of stuck in Pset7 problem 3b. It's about finding the centroid of an isosceles triangle for any angle theta between the equal sides. The method I'm trying is to use polar coordinates and I split the triangle through its height line and compute firstly the outer integrals from 0 to theta/2 and then from theta to theta/2 and then add them up together. I set the limits of the inner integral from 0 to R*cos(theta/2), where R is the length of the equal sides. After computing this it gives something like cos^4(theta/2)*sin(theta/2) which I'd

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Hi, I am new to this forum. Greetings to everyone. I'm kind of stuck in Pset7 problem 3b. It's about finding the centroid of an isosceles triangle for any angle theta between the equal sides. The method I'm trying is to use polar coordinates and I split the triangle through its height line and compute firstly the outer integrals from 0 to theta/2 and then from theta to theta/2 and then add them up together. I set the limits of the inner integral from 0 to R*cos(theta/2), where R is the length of the equal sides. After computing this it gives something like cos^4(theta/2)*sin(theta/2) which I'd

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why are u using polar coordinates? You can just use basic euclidean geometry to find the centroid: if it is equilateral, say side length is 2a, so u wont have to deal with as many fractions so, as median is equal to altitude,which is a*rt(3), and 2/3 of median from any vertex takes u to centroid; centroid is the coordinates of any one of the vertices, so say it is (x, y), plus 2/3 of median, multiplied by correct theta, found as follows: so, we want 2/3 of median, also altitude. As it is altitude, it divides the equilateral triangle into two 30-60-90 triangles, and the angle between vertex and adjacent side is 30 degrees, or pi/6 rad. As such, the x-component added to vertex x-component, to reach centroid, is 2/3(altitude)*cos(30)=2a/rt(3) *cos(30)=a Similarly, the y-component added to vertex y-component, to reach centroid, is 2/3(altitude)*sin(30)=2a/rt(3) *sin(30)=2/3 *a So, centroid coordinates: (assuming one vertex coordinate is (x,y) or w/e u want to call it) (x+a, y+ 2/3 *a) I haven't actually looked at the problem, which is why im using variables instead of numbers (if numbers are given), so just plug in the numbers you were given. This formula should work for any equilateral triangle.

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