Does the average height change after we squished the base?
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My question is, for an arbitrary surface $z = f(x,y)$, if the base coordinate x and y has experienced some linear transformation. Can I multiply the old calculated average height in Z direction to the new base area in XY to get the new volume?
solid-geometry
asked Aug 23 at 2:44
Zhidong Li
224
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My question is, for an arbitrary surface $z = f(x,y)$, if the base coordinate x and y has experienced some linear transformation. Can I multiply the old calculated average height in Z direction to the new base area in XY to get the new volume?
solid-geometry
asked Aug 23 at 2:44
Zhidong Li
224
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
My question is, for an arbitrary surface $z = f(x,y)$, if the base coordinate x and y has experienced some linear transformation. Can I multiply the old calculated average height in Z direction to the new base area in XY to get the new volume?
solid-geometry
asked Aug 23 at 2:44
Zhidong Li
224
My question is, for an arbitrary surface $z = f(x,y)$, if the base coordinate x and y has experienced some linear transformation. Can I multiply the old calculated average height in Z direction to the new base area in XY to get the new volume?
solid-geometry
asked Aug 23 at 2:44
Zhidong Li
224
edited Aug 23 at 2:54
asked Aug 23 at 2:44
Zhidong Li
224
asked Aug 23 at 2:44
Zhidong Li
224
asked Aug 23 at 2:44
Zhidong Li
224
224
add a comment |Â
add a comment |Â
1 Answer
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Yes if you update $f(x,y)$ so it gives the same height over the transformed point as the old $f(x,y)$ did. The average height is $$overline h=frac int f(x,y)dxdyint dxdy$$ where $int dxdy$ is the area of the base. If you apply a linear transformation to get to $(u,v)$ the area of the base will be $int dudv$, which will be the old area times the Jacobian. Each small slice of the volume will have its area multiplied by the same amount.
answered Aug 23 at 3:18
Ross Millikan
278k22188355
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes if you update $f(x,y)$ so it gives the same height over the transformed point as the old $f(x,y)$ did. The average height is $$overline h=frac int f(x,y)dxdyint dxdy$$ where $int dxdy$ is the area of the base. If you apply a linear transformation to get to $(u,v)$ the area of the base will be $int dudv$, which will be the old area times the Jacobian. Each small slice of the volume will have its area multiplied by the same amount.
answered Aug 23 at 3:18
Ross Millikan
278k22188355
add a comment |Â
up vote
1
down vote
accepted
Yes if you update $f(x,y)$ so it gives the same height over the transformed point as the old $f(x,y)$ did. The average height is $$overline h=frac int f(x,y)dxdyint dxdy$$ where $int dxdy$ is the area of the base. If you apply a linear transformation to get to $(u,v)$ the area of the base will be $int dudv$, which will be the old area times the Jacobian. Each small slice of the volume will have its area multiplied by the same amount.
answered Aug 23 at 3:18
Ross Millikan
278k22188355
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes if you update $f(x,y)$ so it gives the same height over the transformed point as the old $f(x,y)$ did. The average height is $$overline h=frac int f(x,y)dxdyint dxdy$$ where $int dxdy$ is the area of the base. If you apply a linear transformation to get to $(u,v)$ the area of the base will be $int dudv$, which will be the old area times the Jacobian. Each small slice of the volume will have its area multiplied by the same amount.
answered Aug 23 at 3:18
Ross Millikan
278k22188355
Yes if you update $f(x,y)$ so it gives the same height over the transformed point as the old $f(x,y)$ did. The average height is $$overline h=frac int f(x,y)dxdyint dxdy$$ where $int dxdy$ is the area of the base. If you apply a linear transformation to get to $(u,v)$ the area of the base will be $int dudv$, which will be the old area times the Jacobian. Each small slice of the volume will have its area multiplied by the same amount.
answered Aug 23 at 3:18
Ross Millikan
278k22188355
answered Aug 23 at 3:18
Ross Millikan
278k22188355
answered Aug 23 at 3:18
Ross Millikan
278k22188355
answered Aug 23 at 3:18
Ross Millikan
278k22188355
278k22188355
add a comment |Â
add a comment |Â
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