Integral of a 2-Form Over a Certain Region of Integration
This is a rather simple problem, but one that I'm struggling with nonetheless. I'm given a 2-Form, $beta = zdx wedge dy-x^2dy wedge dz$ that I need to integrate over the surface S : $z=4-x^2-y^2 ge 0 $.
Parametrising the surface, I have:
$G(x,y) = < x,y,4-x^2-y^2>$ and the Jacobian: $$ begin{bmatrix} DG end{bmatrix} = begin {bmatrix}1 & 0 \0 & 1 \ -2x & -2y end{bmatrix}$$
Evaluating the 2-Form, I have $int_Sbeta= iint_{z=4-x^2-y^2} (4-x^2-y^2) begin{vmatrix}1&0 \ 0&1 end {vmatrix} - x^2 begin{vmatrix} 0 & 1\ -2x & -2y end{vmatrix}dxdy$
This simplifies to: $int_Sbeta = iint_{z=4-x^2-y^2} 4 -x^2-y^2-2x^3dxdy$
In determining the region of integration, though, I get stuck. My gut is telling me to go with polar coordinates. If this were the case, I know $0 le theta le 2{pi}$, but I'm uncertain of the limits of integration for $r$. Would it be $0 le r le 2$? I don't think this is correct but am uncertain of any other way you'd do it.
calculus integration multivariable-calculus polar-coordinates vector-fields
asked Dec 1 at 0:14
Jackson Joffe
575
add a comment |
This is a rather simple problem, but one that I'm struggling with nonetheless. I'm given a 2-Form, $beta = zdx wedge dy-x^2dy wedge dz$ that I need to integrate over the surface S : $z=4-x^2-y^2 ge 0 $.
Parametrising the surface, I have:
$G(x,y) = < x,y,4-x^2-y^2>$ and the Jacobian: $$ begin{bmatrix} DG end{bmatrix} = begin {bmatrix}1 & 0 \0 & 1 \ -2x & -2y end{bmatrix}$$
Evaluating the 2-Form, I have $int_Sbeta= iint_{z=4-x^2-y^2} (4-x^2-y^2) begin{vmatrix}1&0 \ 0&1 end {vmatrix} - x^2 begin{vmatrix} 0 & 1\ -2x & -2y end{vmatrix}dxdy$
This simplifies to: $int_Sbeta = iint_{z=4-x^2-y^2} 4 -x^2-y^2-2x^3dxdy$
In determining the region of integration, though, I get stuck. My gut is telling me to go with polar coordinates. If this were the case, I know $0 le theta le 2{pi}$, but I'm uncertain of the limits of integration for $r$. Would it be $0 le r le 2$? I don't think this is correct but am uncertain of any other way you'd do it.
calculus integration multivariable-calculus polar-coordinates vector-fields
asked Dec 1 at 0:14
Jackson Joffe
575
Your intuition is correct - if you want to make sure, just plot all the points in the integration region, and you will recognize it is a circle of radius 2, so the above parameterization would work.
– Neeyanth Kopparapu
Dec 1 at 1:35
So basically this is saying that r spans from 0 to 2. But here, my region of integration is a paraboloid. Say my region of integration was a circle of radius 2. The limits of integration would be the same as they are for my paraboloid, but they are clearly different shapes.
– Jackson Joffe
Dec 1 at 18:21
add a comment |
This is a rather simple problem, but one that I'm struggling with nonetheless. I'm given a 2-Form, $beta = zdx wedge dy-x^2dy wedge dz$ that I need to integrate over the surface S : $z=4-x^2-y^2 ge 0 $.
Parametrising the surface, I have:
$G(x,y) = < x,y,4-x^2-y^2>$ and the Jacobian: $$ begin{bmatrix} DG end{bmatrix} = begin {bmatrix}1 & 0 \0 & 1 \ -2x & -2y end{bmatrix}$$
Evaluating the 2-Form, I have $int_Sbeta= iint_{z=4-x^2-y^2} (4-x^2-y^2) begin{vmatrix}1&0 \ 0&1 end {vmatrix} - x^2 begin{vmatrix} 0 & 1\ -2x & -2y end{vmatrix}dxdy$
This simplifies to: $int_Sbeta = iint_{z=4-x^2-y^2} 4 -x^2-y^2-2x^3dxdy$
In determining the region of integration, though, I get stuck. My gut is telling me to go with polar coordinates. If this were the case, I know $0 le theta le 2{pi}$, but I'm uncertain of the limits of integration for $r$. Would it be $0 le r le 2$? I don't think this is correct but am uncertain of any other way you'd do it.
calculus integration multivariable-calculus polar-coordinates vector-fields
asked Dec 1 at 0:14
Jackson Joffe
575
This is a rather simple problem, but one that I'm struggling with nonetheless. I'm given a 2-Form, $beta = zdx wedge dy-x^2dy wedge dz$ that I need to integrate over the surface S : $z=4-x^2-y^2 ge 0 $.
Parametrising the surface, I have:
$G(x,y) = < x,y,4-x^2-y^2>$ and the Jacobian: $$ begin{bmatrix} DG end{bmatrix} = begin {bmatrix}1 & 0 \0 & 1 \ -2x & -2y end{bmatrix}$$
Evaluating the 2-Form, I have $int_Sbeta= iint_{z=4-x^2-y^2} (4-x^2-y^2) begin{vmatrix}1&0 \ 0&1 end {vmatrix} - x^2 begin{vmatrix} 0 & 1\ -2x & -2y end{vmatrix}dxdy$
This simplifies to: $int_Sbeta = iint_{z=4-x^2-y^2} 4 -x^2-y^2-2x^3dxdy$
In determining the region of integration, though, I get stuck. My gut is telling me to go with polar coordinates. If this were the case, I know $0 le theta le 2{pi}$, but I'm uncertain of the limits of integration for $r$. Would it be $0 le r le 2$? I don't think this is correct but am uncertain of any other way you'd do it.
calculus integration multivariable-calculus polar-coordinates vector-fields
calculus integration multivariable-calculus polar-coordinates vector-fields
asked Dec 1 at 0:14
Jackson Joffe
575
asked Dec 1 at 0:14
Jackson Joffe
575
asked Dec 1 at 0:14
Jackson Joffe
575
asked Dec 1 at 0:14
Jackson Joffe
575
asked Dec 1 at 0:14
Jackson Joffe
575
575
Your intuition is correct - if you want to make sure, just plot all the points in the integration region, and you will recognize it is a circle of radius 2, so the above parameterization would work.
– Neeyanth Kopparapu
Dec 1 at 1:35
So basically this is saying that r spans from 0 to 2. But here, my region of integration is a paraboloid. Say my region of integration was a circle of radius 2. The limits of integration would be the same as they are for my paraboloid, but they are clearly different shapes.
– Jackson Joffe
Dec 1 at 18:21
add a comment |
Your intuition is correct - if you want to make sure, just plot all the points in the integration region, and you will recognize it is a circle of radius 2, so the above parameterization would work.
– Neeyanth Kopparapu
Dec 1 at 1:35
So basically this is saying that r spans from 0 to 2. But here, my region of integration is a paraboloid. Say my region of integration was a circle of radius 2. The limits of integration would be the same as they are for my paraboloid, but they are clearly different shapes.
– Jackson Joffe
Dec 1 at 18:21
Your intuition is correct - if you want to make sure, just plot all the points in the integration region, and you will recognize it is a circle of radius 2, so the above parameterization would work.
– Neeyanth Kopparapu
Dec 1 at 1:35
Your intuition is correct - if you want to make sure, just plot all the points in the integration region, and you will recognize it is a circle of radius 2, so the above parameterization would work.
– Neeyanth Kopparapu
Dec 1 at 1:35
So basically this is saying that r spans from 0 to 2. But here, my region of integration is a paraboloid. Say my region of integration was a circle of radius 2. The limits of integration would be the same as they are for my paraboloid, but they are clearly different shapes.
– Jackson Joffe
Dec 1 at 18:21
So basically this is saying that r spans from 0 to 2. But here, my region of integration is a paraboloid. Say my region of integration was a circle of radius 2. The limits of integration would be the same as they are for my paraboloid, but they are clearly different shapes.
– Jackson Joffe
Dec 1 at 18:21
add a comment |
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Your intuition is correct - if you want to make sure, just plot all the points in the integration region, and you will recognize it is a circle of radius 2, so the above parameterization would work.
– Neeyanth Kopparapu
Dec 1 at 1:35
So basically this is saying that r spans from 0 to 2. But here, my region of integration is a paraboloid. Say my region of integration was a circle of radius 2. The limits of integration would be the same as they are for my paraboloid, but they are clearly different shapes.
– Jackson Joffe
Dec 1 at 18:21