 My Math Forum Unoriented Surface Integral- Calculus III

 Calculus Calculus Math Forum

 April 21st, 2017, 07:19 AM #1 Newbie   Joined: Nov 2016 From: Louisville, Ky Posts: 7 Thanks: 0 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! April 21st, 2017, 02:59 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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. April 23rd, 2017, 08:03 AM   #3
Newbie

Joined: Nov 2016
From: Louisville, Ky

Posts: 7
Thanks: 0

Quote:
 Originally Posted by Country Boy 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! April 23rd, 2017, 09:03 AM   #4
Newbie

Joined: Nov 2016
From: Louisville, Ky

Posts: 7
Thanks: 0

Quote:
 Originally Posted by Country Boy 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) Tags calculus, iii, integral, surface, unoriented ### unoriented integral

Click on a term to search for related topics.
 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post SirSam Calculus 3 February 23rd, 2017 09:02 AM henrymerrild Calculus 2 May 1st, 2014 11:33 AM Jhenrique Calculus 2 December 27th, 2013 08:14 PM kriko Calculus 1 August 21st, 2010 11:55 AM sansar Calculus 1 May 10th, 2009 05:21 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      