Assassin315
New member
- Joined
- Oct 26, 2010
- Messages
- 12
Huge multi-step problem here involving centroids. I looked up the basics of how they work, but I'm having difficulties starting this.
I just got done with multiple surface area of revolution problems, and this is the last part of the homework.
The problem opens by stating "The point (x-bar, y-bar) is called the centroid of the curve if its coordinates are defined by:"
A) If the curve is said to be uniform, its centroid is its center of gravity. Determine the centroid of semicircle x^2 + y^2 = a^2, y >= 0.
B) In part A, let the curve be revolved around the x-axis. Show that the area of the surface of revolution equals the curve length times the distance traveled by the centroid on its trip around the x-axis.
C) Use the theorem proved in part B to find:
1) y-bar for the semicircular arc in part A.
2) the surface area of the torus (a doughnut) obtained by revolving the circle x^2 + (y-b)^2 = a^2, a =< b, about the x-axis.
Alright, so I can't do B and C without doing A, so I tried solving A... and failed. I got an x term (sqrt(a^2 + y^2)) and a y term (sqrt(a^2 + x^2)). I assume to find x-bar and y-bar, I take the integrals of x and y, find ds (how -- use something like sqrt(1+ (dy/dx)^2)?), and use something like sqrt(x^2 + y^2) as my curve length? Maybe? I'm kinda tired, so I'll try this more tomorrow, but I could use a few hints.
I just got done with multiple surface area of revolution problems, and this is the last part of the homework.
The problem opens by stating "The point (x-bar, y-bar) is called the centroid of the curve if its coordinates are defined by:"
A) If the curve is said to be uniform, its centroid is its center of gravity. Determine the centroid of semicircle x^2 + y^2 = a^2, y >= 0.
B) In part A, let the curve be revolved around the x-axis. Show that the area of the surface of revolution equals the curve length times the distance traveled by the centroid on its trip around the x-axis.
C) Use the theorem proved in part B to find:
1) y-bar for the semicircular arc in part A.
2) the surface area of the torus (a doughnut) obtained by revolving the circle x^2 + (y-b)^2 = a^2, a =< b, about the x-axis.
Alright, so I can't do B and C without doing A, so I tried solving A... and failed. I got an x term (sqrt(a^2 + y^2)) and a y term (sqrt(a^2 + x^2)). I assume to find x-bar and y-bar, I take the integrals of x and y, find ds (how -- use something like sqrt(1+ (dy/dx)^2)?), and use something like sqrt(x^2 + y^2) as my curve length? Maybe? I'm kinda tired, so I'll try this more tomorrow, but I could use a few hints.