My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree2Thanks
  • 2 Post By Country Boy
Reply
 
LinkBack Thread Tools Display Modes
April 21st, 2017, 07:19 AM   #1
Newbie
 
Joined: Nov 2016
From: Louisville, Ky

Posts: 7
Thanks: 0

Post Unoriented Surface Integral- Calculus III

Hello, I need help with this problem!
- Use the unoriented surface integral to express the mass of a dome in the shape of the paraboloid z= 25-x^2-y^2 , that lies above the xy-plane with density function f(x,y,z)=x^2+y^2 as a double integral over a suitable region D in the xy-plane. Evaluate the integral.



Thank you!
keishajc is offline  
 
April 21st, 2017, 02:59 PM   #2
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 2,822
Thanks: 750

The mass of a two-dimensional object, S, given a density function f(x,y), is just the integral of that density function over S: where "dS" is the "differential of surface area" in terms of x and y.

One way to do that is to find dS is first to find the vector differential, . Then then scalar differential of surface area is the magnitude of that vector: .

If a surface is given in parametric equations, x= f(u, v), y= g(u, v), z= h(u, v) then the two vectors and lie in the tangent plane and their cross product, is perpendicular to the surface and is so its length is dS.

Here, you are given the surface as z= f(x, y) so you can take x and y as the parameters u and v: x= u, y= v, z= f(u, v). Then the vector derivative with respect to u is , the vector derivative with respect to v is and their cross product is and .
Thanks from greg1313 and keishajc

Last edited by greg1313; April 21st, 2017 at 03:13 PM.
Country Boy is online now  
April 23rd, 2017, 08:03 AM   #3
Newbie
 
Joined: Nov 2016
From: Louisville, Ky

Posts: 7
Thanks: 0

Quote:
Originally Posted by Country Boy View Post
The mass of a two-dimensional object, S, given a density function f(x,y), is just the integral of that density function over S: where "dS" is the "differential of surface area" in terms of x and y.

One way to do that is to find dS is first to find the vector differential, . Then then scalar differential of surface area is the magnitude of that vector: .

If a surface is given in parametric equations, x= f(u, v), y= g(u, v), z= h(u, v) then the two vectors and lie in the tangent plane and their cross product, is perpendicular to the surface and is so its length is dS.

Here, you are given the surface as z= f(x, y) so you can take x and y as the parameters u and v: x= u, y= v, z= f(u, v). Then the vector derivative with respect to u is , the vector derivative with respect to v is and their cross product is and .
I understand how to set it up but it is the limits of integration and the actual integral that are bothering me! Thank you!
keishajc is offline  
April 23rd, 2017, 09:03 AM   #4
Newbie
 
Joined: Nov 2016
From: Louisville, Ky

Posts: 7
Thanks: 0

Quote:
Originally Posted by Country Boy View Post
The mass of a two-dimensional object, S, given a density function f(x,y), is just the integral of that density function over S: where "dS" is the "differential of surface area" in terms of x and y.

One way to do that is to find dS is first to find the vector differential, . Then then scalar differential of surface area is the magnitude of that vector: .

If a surface is given in parametric equations, x= f(u, v), y= g(u, v), z= h(u, v) then the two vectors and lie in the tangent plane and their cross product, is perpendicular to the surface and is so its length is dS.

Here, you are given the surface as z= f(x, y) so you can take x and y as the parameters u and v: x= u, y= v, z= f(u, v). Then the vector derivative with respect to u is , the vector derivative with respect to v is and their cross product is and .
OKay, update! I have made it and gotten the integral from 0-2pi and 0-5 of r^2(sqrt(4r^2+1)) dr d(theta)
keishajc is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
calculus, iii, integral, surface, unoriented



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Calculus (Surface Area of Solid Revolution) SirSam Calculus 3 February 23rd, 2017 09:02 AM
What is a surface integral? henrymerrild Calculus 2 May 1st, 2014 11:33 AM
Fundamental theorem of calculus for surface integrals? Jhenrique Calculus 2 December 27th, 2013 08:14 PM
surface integral kriko Calculus 1 August 21st, 2010 11:55 AM
Surface Integral sansar Calculus 1 May 10th, 2009 05:21 AM





Copyright © 2017 My Math Forum. All rights reserved.