Surface integral over graph of a function
$begingroup$
Q: Evaluate the surface integral $iint_S mathbf{F}cdot dmathbf{S}$, where $mathbf{F}(x,y,z)= 3x^2mathbf{i}-2xymathbf{j}+8mathbf{k}$, and $S$ is the graph of the function $z=f(x,y)=2x-y$ for $0leq xleq2$ and $0leq y leq2$.
Attempt: This question is from a previous years Vector Calculus exam. When attempting to solve, I had thought to parameterize the surface, and then use $int_a^b mathbf{F}(mathbf{c}(t))cdot mathbf{c}'(t),dt$, $a leq t leq b$. In the solutions, they have $$iint_S mathbf{F}cdot dmathbf{S} = int_0^2int_0^2 (-6x^2-2xy+8),dxdy= -8$$
I'm mostly looking for an explanation as to how they arrived at the expression inside the integral ($-6x^2-2xy+8$), any help would be greatly appreciated.
Thanks.
calculus multivariable-calculus
edited Dec 26 '18 at 13:06
mechanodroid
27.6k62447
asked Jun 13 '13 at 7:19
user68093user68093
603
$endgroup$
add a comment |
$begingroup$
Q: Evaluate the surface integral $iint_S mathbf{F}cdot dmathbf{S}$, where $mathbf{F}(x,y,z)= 3x^2mathbf{i}-2xymathbf{j}+8mathbf{k}$, and $S$ is the graph of the function $z=f(x,y)=2x-y$ for $0leq xleq2$ and $0leq y leq2$.
Attempt: This question is from a previous years Vector Calculus exam. When attempting to solve, I had thought to parameterize the surface, and then use $int_a^b mathbf{F}(mathbf{c}(t))cdot mathbf{c}'(t),dt$, $a leq t leq b$. In the solutions, they have $$iint_S mathbf{F}cdot dmathbf{S} = int_0^2int_0^2 (-6x^2-2xy+8),dxdy= -8$$
I'm mostly looking for an explanation as to how they arrived at the expression inside the integral ($-6x^2-2xy+8$), any help would be greatly appreciated.
Thanks.
calculus multivariable-calculus
edited Dec 26 '18 at 13:06
mechanodroid
27.6k62447
asked Jun 13 '13 at 7:19
user68093user68093
603
$endgroup$
add a comment |
$begingroup$
Q: Evaluate the surface integral $iint_S mathbf{F}cdot dmathbf{S}$, where $mathbf{F}(x,y,z)= 3x^2mathbf{i}-2xymathbf{j}+8mathbf{k}$, and $S$ is the graph of the function $z=f(x,y)=2x-y$ for $0leq xleq2$ and $0leq y leq2$.
Attempt: This question is from a previous years Vector Calculus exam. When attempting to solve, I had thought to parameterize the surface, and then use $int_a^b mathbf{F}(mathbf{c}(t))cdot mathbf{c}'(t),dt$, $a leq t leq b$. In the solutions, they have $$iint_S mathbf{F}cdot dmathbf{S} = int_0^2int_0^2 (-6x^2-2xy+8),dxdy= -8$$
I'm mostly looking for an explanation as to how they arrived at the expression inside the integral ($-6x^2-2xy+8$), any help would be greatly appreciated.
Thanks.
calculus multivariable-calculus
edited Dec 26 '18 at 13:06
mechanodroid
27.6k62447
asked Jun 13 '13 at 7:19
user68093user68093
603
$endgroup$
Q: Evaluate the surface integral $iint_S mathbf{F}cdot dmathbf{S}$, where $mathbf{F}(x,y,z)= 3x^2mathbf{i}-2xymathbf{j}+8mathbf{k}$, and $S$ is the graph of the function $z=f(x,y)=2x-y$ for $0leq xleq2$ and $0leq y leq2$.
Attempt: This question is from a previous years Vector Calculus exam. When attempting to solve, I had thought to parameterize the surface, and then use $int_a^b mathbf{F}(mathbf{c}(t))cdot mathbf{c}'(t),dt$, $a leq t leq b$. In the solutions, they have $$iint_S mathbf{F}cdot dmathbf{S} = int_0^2int_0^2 (-6x^2-2xy+8),dxdy= -8$$
I'm mostly looking for an explanation as to how they arrived at the expression inside the integral ($-6x^2-2xy+8$), any help would be greatly appreciated.
Thanks.
calculus multivariable-calculus
calculus multivariable-calculus
edited Dec 26 '18 at 13:06
mechanodroid
27.6k62447
asked Jun 13 '13 at 7:19
user68093user68093
603
edited Dec 26 '18 at 13:06
mechanodroid
27.6k62447
asked Jun 13 '13 at 7:19
user68093user68093
603
edited Dec 26 '18 at 13:06
mechanodroid
27.6k62447
edited Dec 26 '18 at 13:06
mechanodroid
27.6k62447
edited Dec 26 '18 at 13:06
mechanodroid
27.6k62447
27.6k62447
asked Jun 13 '13 at 7:19
user68093user68093
603
asked Jun 13 '13 at 7:19
user68093user68093
603
asked Jun 13 '13 at 7:19
user68093user68093
603
603
add a comment |
add a comment |
1 Answer
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$begingroup$
The surface integral is better written as
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$
where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is
$$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$
Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:
$$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$
Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$
which is the result you wanted to see.
answered Jun 13 '13 at 8:16
Ron GordonRon Gordon
122k14155265
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
The surface integral is better written as
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$
where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is
$$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$
Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:
$$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$
Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$
which is the result you wanted to see.
answered Jun 13 '13 at 8:16
Ron GordonRon Gordon
122k14155265
$endgroup$
add a comment |
$begingroup$
The surface integral is better written as
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$
where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is
$$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$
Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:
$$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$
Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$
which is the result you wanted to see.
answered Jun 13 '13 at 8:16
Ron GordonRon Gordon
122k14155265
$endgroup$
add a comment |
$begingroup$
The surface integral is better written as
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$
where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is
$$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$
Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:
$$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$
Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$
which is the result you wanted to see.
answered Jun 13 '13 at 8:16
Ron GordonRon Gordon
122k14155265
$endgroup$
The surface integral is better written as
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F}$$
where $hat{mathbf{n}}$ is the unit surface normal. For surfaces of the form $z=f(x,y)$, the unit normal is
$$hat{mathbf{n}} = frac{displaystyleleft(-frac{partial f}{partial x},-frac{partial f}{partial y},1right)}{displaystylesqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1}}$$
Note that the normalizing factor in the denominator is exactly the stretching factor relating the surface area element to the differential $dx,dy$:
$$dS = sqrt{left(frac{partial f}{partial x}right)^2+left(frac{partial f}{partial y}right)^2+1} , dx, dy$$
Now, $partial f/partial x = 2$ and $partial f/partial y = -1$, so that
$$iint_S dS , hat{mathbf{n}} cdot mathbf{F} = int_0^2 dx , int_0^2 dy , [-(2) cdot (3 x^2) - (-1) cdot (-2 x y) + (1)cdot (8)] $$
which is the result you wanted to see.
answered Jun 13 '13 at 8:16
Ron GordonRon Gordon
122k14155265
answered Jun 13 '13 at 8:16
Ron GordonRon Gordon
122k14155265
answered Jun 13 '13 at 8:16
Ron GordonRon Gordon
122k14155265
answered Jun 13 '13 at 8:16
Ron GordonRon Gordon
122k14155265
122k14155265
add a comment |
add a comment |
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