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 April 21st, 2017, 06:19 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, 01:59 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: $\int\int f(x,y)dS$ 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, $\d\vec{S}$. Then then scalar differential of surface area is the magnitude of that vector: $dS= |d\vec{S}|$. 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, $\times $ is perpendicular to the surface and is $d\vec{S}$ 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 $<1, 0, f_x>$, the vector derivative with respect to v is $<0, 1, f_y>$ and their cross product is $d\vec{S}= =$ and $dS= \sqrt{f_x^2+ f_y^2+ 1}dxdy$. Thanks from greg1313 and keishajc Last edited by greg1313; April 21st, 2017 at 02:13 PM.
April 23rd, 2017, 07:03 AM   #3
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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: $\int\int f(x,y)dS$ 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, $\d\vec{S}$. Then then scalar differential of surface area is the magnitude of that vector: $dS= |d\vec{S}|$. 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, $\times $ is perpendicular to the surface and is $d\vec{S}$ 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 $<1, 0, f_x>$, the vector derivative with respect to v is $<0, 1, f_y>$ and their cross product is $d\vec{S}= =$ and $dS= \sqrt{f_x^2+ f_y^2+ 1}dxdy$.
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, 08:03 AM   #4
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Joined: Nov 2016
From: Louisville, Ky

Posts: 7
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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: $\int\int f(x,y)dS$ 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, $\d\vec{S}$. Then then scalar differential of surface area is the magnitude of that vector: $dS= |d\vec{S}|$. 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, $\times $ is perpendicular to the surface and is $d\vec{S}$ 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 $<1, 0, f_x>$, the vector derivative with respect to v is $<0, 1, f_y>$ and their cross product is $d\vec{S}= =$ and $dS= \sqrt{f_x^2+ f_y^2+ 1}dxdy$.
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)

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