Intergrating a differential form over a triangle
I need to intergrate,
$$int_c x^{1} dx^{2} wedge dx^{3} + x^{2} dx^{3} wedge dx^{4}$$
where $c: E to mathbb R^{4} is$
$$c(u,v) = (uv,u^{2}+v^{2},u-v,log(u+v+1)) = (x^{1}, x^{2}, x^{3}, x^{4})$$
and E is the triangle in the (u,v)-plane with vertices (2,0), (0,2), (0,0).
I started by pulling back the function on c, where $w = x^{1} dx^{2} wedge dx^{3} + x^{2} dx^{3} wedge dx^{4}$such that;
$$c^{*}(w) = -u^{2}v + uv^{2} du wedge dv + frac{2(u^{2} + v^{2})}{u+v+1} duwedge dv $$
Which gives,
$$c^{*}(w) = frac{2(uv^{4} - u^{4}v)}{u+v+1} duwedge dv $$
I might be wrong in something above, but I found integrating this to be very difficult. Could anyone help me out with this?
integration multivariable-calculus
asked Nov 25 at 6:09
LexieStark
12
add a comment |
I need to intergrate,
$$int_c x^{1} dx^{2} wedge dx^{3} + x^{2} dx^{3} wedge dx^{4}$$
where $c: E to mathbb R^{4} is$
$$c(u,v) = (uv,u^{2}+v^{2},u-v,log(u+v+1)) = (x^{1}, x^{2}, x^{3}, x^{4})$$
and E is the triangle in the (u,v)-plane with vertices (2,0), (0,2), (0,0).
I started by pulling back the function on c, where $w = x^{1} dx^{2} wedge dx^{3} + x^{2} dx^{3} wedge dx^{4}$such that;
$$c^{*}(w) = -u^{2}v + uv^{2} du wedge dv + frac{2(u^{2} + v^{2})}{u+v+1} duwedge dv $$
Which gives,
$$c^{*}(w) = frac{2(uv^{4} - u^{4}v)}{u+v+1} duwedge dv $$
I might be wrong in something above, but I found integrating this to be very difficult. Could anyone help me out with this?
integration multivariable-calculus
asked Nov 25 at 6:09
LexieStark
12
add a comment |
I need to intergrate,
$$int_c x^{1} dx^{2} wedge dx^{3} + x^{2} dx^{3} wedge dx^{4}$$
where $c: E to mathbb R^{4} is$
$$c(u,v) = (uv,u^{2}+v^{2},u-v,log(u+v+1)) = (x^{1}, x^{2}, x^{3}, x^{4})$$
and E is the triangle in the (u,v)-plane with vertices (2,0), (0,2), (0,0).
I started by pulling back the function on c, where $w = x^{1} dx^{2} wedge dx^{3} + x^{2} dx^{3} wedge dx^{4}$such that;
$$c^{*}(w) = -u^{2}v + uv^{2} du wedge dv + frac{2(u^{2} + v^{2})}{u+v+1} duwedge dv $$
Which gives,
$$c^{*}(w) = frac{2(uv^{4} - u^{4}v)}{u+v+1} duwedge dv $$
I might be wrong in something above, but I found integrating this to be very difficult. Could anyone help me out with this?
integration multivariable-calculus
asked Nov 25 at 6:09
LexieStark
12
I need to intergrate,
$$int_c x^{1} dx^{2} wedge dx^{3} + x^{2} dx^{3} wedge dx^{4}$$
where $c: E to mathbb R^{4} is$
$$c(u,v) = (uv,u^{2}+v^{2},u-v,log(u+v+1)) = (x^{1}, x^{2}, x^{3}, x^{4})$$
and E is the triangle in the (u,v)-plane with vertices (2,0), (0,2), (0,0).
I started by pulling back the function on c, where $w = x^{1} dx^{2} wedge dx^{3} + x^{2} dx^{3} wedge dx^{4}$such that;
$$c^{*}(w) = -u^{2}v + uv^{2} du wedge dv + frac{2(u^{2} + v^{2})}{u+v+1} duwedge dv $$
Which gives,
$$c^{*}(w) = frac{2(uv^{4} - u^{4}v)}{u+v+1} duwedge dv $$
I might be wrong in something above, but I found integrating this to be very difficult. Could anyone help me out with this?
integration multivariable-calculus
integration multivariable-calculus
asked Nov 25 at 6:09
LexieStark
12
asked Nov 25 at 6:09
LexieStark
12
asked Nov 25 at 6:09
LexieStark
12
asked Nov 25 at 6:09
LexieStark
12
asked Nov 25 at 6:09
LexieStark
12
12
add a comment |
add a comment |
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