Double integral

Double integral

Postby magben » Wed Oct 13, 2021 11:29 am

Calculate [tex]\int \int\limits_{S} 2zdS[/tex], where [tex]S[/tex] is the part of the paraboloid [tex]z^{2}+ y^{2}-2[/tex] below the [tex]xy[/tex] plane.


Guys, I don't know how to resolve this issue. I managed to do the following.

[tex]\int \int\limits_{S} 2( x^{2}+ y^{2 }-2) \sqrt{(2x)^{2}+(2y)^{2}+1} dA[/tex]

[tex]\int \int\limits_{S} 2( x^{2}+ y^{2 }-2) \sqrt{4x^{2}+4y^{2}+1} dA[/tex]

Polar Coordinates:

[tex]x=rcos\theta[/tex]
[tex]y=rsin\theta[/tex]

[tex]dA=rd_rd\theta[/tex]


[tex]\int\limits_{0}^{2\pi} \int\limits_{0}^{\sqrt{2}}2( r^{2}-2) \sqrt{4r^{2}-1}rdrd\theta[/tex]

Can anyone help me resolve this issue or present me with a different resolution?
magben
 
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Re: Double integral

Postby Guest » Sun Oct 24, 2021 5:37 pm

[tex]z^2+ y^2- 2[/tex] is NOT a paraboid.

Do you mean [tex]x= z^2+ y^2- 2[/tex]?
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Re: Double integral

Postby Guest » Sun Nov 07, 2021 9:02 pm

You said "below the xy-plane". Did you mean [tex]z = x^2+ y^2- 2[/tex]?

I am really not at all enthused about helping someone who not only copies a problem wrong but can't check back to say if a proposed correction is right!

But since someone else might be interested: Assuming this is actually [tex]z= x^2+ y^2- 2[/tex] then [tex]\frac{\partial z}{\partial x}= 2x[/tex] and [tex]\frac{\partial z}{\partial y}= 2y[/tex] so [tex]dS= \sqrt{4x^2 + 4y^2}dxdy= 2\sqrt{x^2+ y^2}dxdy= 2rdxdy[/tex]. That "r" is because of the "cylindrical symmetry" and tells us it would be smart to convert to cylindrical coordinates. In cylindrical coordinates [tex]x= r cos(\theta)[/tex] and [tex]y= r sin(\theta)[/tex]` so [tex]dx= cos(\theta)dr- r sin(\theta)d\theta[/tex] and [tex]dy= sin(\theta)dr+ r cos(\theta)d\theta[/tex].

So [tex]dxdy= (cos(\theta)dr- r sin(\theta)d\theta)(sin(\theta)dr+ r cos(\theta)d\theta)= cos(\theta)sin(\theta)drdr+ r cos^2(\theta)drd\theta- r sin^2(\theta)d\theta dr- r^2 sin(\theta)cos(\theta)d\theta d\theta[/tex].

But multiplication of differentials is "anti-commutative" so [tex]drdr= d\theta d\theta= 0[/tex] and [tex]drd\theta= -d\theta dr[/tex]. dxy reduces to
[tex]rcos^2(\theta)d\theta dr+ rsin^2(\theta)d\theta dr= rd\theta dr[/tex] and then [tex]dS= 2r dxdy= 2r^2 d\theta dr[/tex].

Now the graph is [tex]z= x^2+ y^2- 2[/tex] and we want the surface area of the part below the xy-plane. The minimum value of z is -2 when x= y= 0 so r= 0 and z is 0 (the xy-plane) when [tex]x^2+ y^2= 2[/tex] so [tex]r= \sqrt{2}[/tex]. Of course with circular symmetry we can take [tex]\theta[/tex] from 0 to [tex]2\pi[/tex].

That means the surface area is given by [tex]2\int_{\theta= 0}^{2\pi}\int_{r= 0}^{\sqrt{2}} r^2 drd\theta[/tex].
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Re: Double integral

Postby Guest » Mon Jan 03, 2022 10:47 am

Since this has been here for over a month with no response, I will complete the integral.

[tex]2\int_{\theta= 0}^{2\pi}\int_{r= 0}^{\sqrt{2}} r^2 dr d\theta[/tex]
[tex]= 4\pi \int_0^{\sqrt{2}} r^2 dr[/tex]
[tex]= \frac{4}{3}\pi\left[r^3\right]_0^{\sqrt{2}}[/tex]
[tex]= \frac{4}{3}\pi \left(2^{3/2}\right)[/tex]
[tex]= \frac{8\sqrt{2}}{3}\pi[/tex].
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Re: Double integral

Postby Guest » Sun Feb 13, 2022 5:39 am

I still dont get this,...
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