Let $A=(0,1)^2$. Let $\alpha:A\to\Bbb R^3$ be given by the equation
$$\alpha(u,v)=(u,v,u^2+v^2+1)$$
Let $Y$ be the image set of $\alpha$. Evaluate the integral over $Y_\alpha$ of the 2-form $x_2dx_2\land dx_3+x_1 x_3 dx_1\land dx_3$.
Can someone please give me a useful hint for solving this?
Thanks in advance!
On
Feel like I have to add some clarification to my comment above, because the integral is definitely not zero; that's a misuse of Stokes' Theorem. You can tell it's not zero because you're integrating a positive function.
Anyway, the ideas that I had in mind were:
1) Realize the image set as a piece of a boundary of a three dimensional simply connected shape ( for example, extend the boundary downwards to the x-y plane and include the bottom ). Then, since the form is closed, the integral of the exterior derivative on the inside is zero, so the integral on the total boundary is zero by Stokes' theorem. Therefore, the integral on the domain in question is the negative of the integral on simpler ( eg. linear ) sides.
2) The differential is closed and the domain is star shaped, so it's exact, and it's not hard to find a primitive. The integral on the surface is then equal to the integral on the one dimensional boundary, which are again pretty simple.
I should have more time to spell these out in a bit, but wanted to clear up the confusion, especially because the approved answer is now incorrect.
Added: For #1, add the plane $(x,y,1)$ and the vertical sides from the surface to that plane. The integral on the plane is zero because there is no $z$ component to the tangent space there. On the edge where $x=0$, the integral is
$$ \int_0^1\int_1^{1 + y^2}y\,dzdy = \frac{1}{4} $$
For $x=1$ the integral is $$ \int_0^1\int_1^{2+y^2}y\,dzdy = \frac{1}{2} + \frac{1}{4} $$
For $y=0$ the integral is $$ \int_0^1\int_1^{1+x^2}xz\,dzdx = \int_0^1(x\cdot\frac{1}{2}((1+x^2)^2 - 1)dx $$
and for $y=1$ the integral is $$ \int_0^1\int_1^{2+x^2}xz\,dzdx = \int_0^1(x\cdot\frac{1}{2}((2+x^2)^2 -1)dx $$
So, the total surface integral is the top minus the bottom plus the $y=0$ and $x=1$ sides and minus the $y=1$ and $x=0$ sides ( by orientation ).
The $y=0$ minus the $y=1$ side is
$$ \frac{1}{2}\int_0^{1}x(((1+x^2)^2 - 1) - ((1 + 1 + x^2)^2 - 1 )dx = -\frac{1}{2}\int_0^1x(3+2x^2)dx = -1 $$
and $x=1$ minus the $x=0$ sides is $\frac{1}{2}$. So since the total surface integral is zero thanks to stokes theorem, the integral we're looking for is negative the sum of the rest, so $\frac{1}{2}$
Will add number 2 as well.
Stokes' Method 2) It's not too hard to see that your differential is $d(-yzdy + \frac{1}{2}(x^2z)dz)$, so by Stokes' Theorem, the integral we're looking for is the integral of that primitive on the boundary.
The boundary has four components. For $y=0$ component, the integral is $$ \int_0^1\frac{1}{2}(x^2\cdot(1+x^2)\cdot 2x)dx $$
For $y=1$ it is $$ \int_0^1\frac{1}{2}(x^2\cdot(2+x^2)\cdot 2x)dx $$
For $x=0$ it is $$ \int_0^1-(y\cdot(1+y^2))dy $$
And for $x=1$ it is $$ \int_0^1\left(-(y\cdot(2+y^2)) + \frac{1}{2}(2+y^2)2y\right)dy $$
With the given orientation, we need to add the $y=0$ and $x=1$ integrals and subtract the $y=1$ and $x=0$ integrals. $y=0$ minus $y=1$ is
$$ \int_0^1 -x^3dx = -\frac{1}{4} $$
The $x=1$ integral is $0$ so and the $x=0$ integral is $\frac{3}{4}$. This adds up to $\frac{1}{2}$, agreeing with the other method, as well as the method suggested by @Frieder
Given parametrization: $$\phi (u,v) = (u,v,{u^2} + {v^2} + 1)$$ Setting $$x = u,y = v,z = {u^2} + {v^2} + 1$$ Then calculate differential resp. to parametrization: $$\begin{gathered} dx = du \hfill \\ dy = dv \hfill \\ dz = 2udu + 2vdv \hfill \\ \end{gathered} $$ Form in old coordinates: $$\omega (x,y,z) = ydy \wedge dz + xzdx \wedge dz = (ydy + xzdx) \wedge dz$$ Form in new coordinates resp. surface, the world famous pull-back: $$\begin{gathered} {\phi ^ * }\omega (u,v) = (u({u^2} + {v^2} + 1)du + vdv) \wedge (2udu + 2vdv) \hfill \\ {\phi ^ * }\omega (u,v) = 2uv({u^2} + {v^2} + 1)du \wedge dv - 2uvdu \wedge dv \hfill \\ {\phi ^ * }\omega (u,v) = 2({u^3}v + u{v^3})du \wedge dv \hfill \\ \end{gathered}$$ Ready to integrate:$$2 \cdot \int\limits_0^1 {\int\limits_0^1 {({u^3}v + u{v^3})dudv} } = \int\limits_0^1 {(\frac{1}{2}v + {v^3})dv = } \frac{1}{2}$$
Another user (Callus) saw this:
$$\begin{gathered} d\omega = d(xzdx \wedge dz) \hfill \\ = d(xz) \wedge dx \wedge dz \hfill \\ = (zdx + xdz) \wedge dx \wedge dz \hfill \\ = zdx \wedge dx \wedge dz + xdz \wedge dx \wedge dz = 0 \hfill \\ \end{gathered} $$
That means the form is closed. On the other hand: We know formula for exterior derivative for wedge of forms: $$d(\alpha \wedge \beta ) = d\alpha \wedge \beta + {( - 1)^k}\alpha \wedge d\beta$$ Here $\alpha$ is a k-form. Therefore $$\omega = y{\kern 1pt} dy \wedge dz + xz{\kern 1pt} dx \wedge dz$$ may be written like: $$\omega = d(\frac{1}{2}{y^2} \wedge dz) + d(\frac{1}{2}{x^2} \wedge zdz) = d(\frac{1}{2}({y^2}dz + {x^2}z)dz)$$ Or, if we set: $$\theta = \frac{1}{2}({x^2}z + {y^2})dz$$ we get for short: $$\omega = d\theta $$ At least resp. local coordinates, $\omega$ is exact. It's the exterior derivative from another form $\theta$. So, by Stoke's theorem, that means: $$\int\limits_A {d\theta } = \int\limits_{\partial A} \theta = \int\limits_{\partial A} {\frac{1}{2}({x^2}z + {y^2})dz}$$ Let's evaluate the RHS: If $z = {x^2} + {y^2} + 1$, so $dz = 2xdx + 2ydy$ and for RHS we get: $$\int\limits_{\partial A} {\frac{1}{2}({x^2}z + {y^2})dz} = \int\limits_{\partial A} {({x^2}({x^2} + {y^2} + 1) + {y^2})xdx} + \int\limits_{\partial A} {({x^2}({x^2} + {y^2} + 1) + {y^2})ydy}$$ With counterclockwise orientation of boundary $\partial A = A\backslash (0,1) \times (0,1)$, we have to manage following four integrals: $$\left. {\begin{array}{*{20}{c}} {x \in [0,1]} \\ {y = 0} \end{array}} \right\}\int\limits_0^1 {({x^5} + {x^3})dx} = \frac{5}{{12}}$$
$$\left. {\begin{array}{*{20}{c}} {x = 1} \\ {y \in [0,1]} \end{array}} \right\}2\int\limits_0^1 {({y^3} + y)dy} = \frac{3}{2}$$
$$\left. {\begin{array}{*{20}{c}} {x \in [0,1]} \\ {y = 1} \end{array}} \right\} - \int\limits_0^1 {({x^5} + 2{x^3} + x)dx} = - \frac{1}{6} - \frac{1}{2} - \frac{1}{2} = - \frac{7}{6}$$
$$\left. {\begin{array}{*{20}{c}} {x = 0} \\ {y \in [0,1]} \end{array}} \right\} - \int\limits_0^1 {{y^3}dy} = - \frac{1}{4}$$
Last two integrals got minus-sign resp. orientation. Summing up, we have: $$\int\limits_{\partial A} {\frac{1}{2}({x^2}z + {y^2})dz} = \frac{5}{{12}} + \frac{3}{2} - \frac{{14}}{{12}} - \frac{3}{{12}} = \frac{6}{{12}} = \frac{1}{2}$$