Evaluating Surface Integral for Hyperboloid in Cylinder

[bugatti79]

Homework Statement

Evaluate the surface integral

Homework Equations

f(x,y,z)=y where sigma is part of the hyperboloid y=x^2+z^2 that lies inside cylinder x^2+z^2=4

The Attempt at a Solution

For [itex] \displaystyle S= \int \int_R \sqrt(z_x^2+z_y^2+1) dA[/itex]

I calculate [itex]\displaystyle \sqrt(z_x^2+z_y^2+1)=\sqrt( \frac{x^2}{\sqrt{y-x^2}}+ \frac{1}{\sqrt{y-x^2}}+1)[/itex]
I have tried simplifying this further but it still looks ugly...any suggestions on how to continue and evalute the surface integral?

Thanks

 

[Silversonic]

Homework Statement

Evaluate the surface integral

Homework Equations

f(x,y,z)=y where sigma is part of the hyperboloid y=x^2+z^2 that lies inside cylinder x^2+z^2=4

The Attempt at a Solution

For [itex] \displaystyle S= \int \int_R \sqrt(z_x^2+z_y^2+1) dA[/itex]

I calculate [itex]\displaystyle \sqrt(z_x^2+z_y^2+1)=\sqrt( \frac{x^2}{\sqrt{y-x^2}}+ \frac{1}{\sqrt{y-x^2}}+1)[/itex]
I have tried simplifying this further but it still looks ugly...any suggestions on how to continue and evalute the surface integral?

Thanks

I assume you've projected on to the xz-plane, after all that would be the easier option.

It may be easier to stick with the explicit equation already given for y. So you attain the normal vector

[itex] n = \frac{(2x,-1,2z)}{\sqrt{4x^2 + 4z^2 + 1}} [/itex]

Projecting on to the xz-plane we get;

[itex] |n.j| = \frac{1}{\sqrt{4x^2 + 4z^2 + 1}} [/itex]

So

[itex] dS = \frac{dA_y}{|n.j|} = \sqrt{4x^2 + 4z^2 + 1} dA_y [/itex]

After that it would be ideal to convert to polar co-ordinates. You also seem to have left the original y out of the surface integral.

 

[bugatti79]

I assume you've projected on to the xz-plane, after all that would be the easier option.

It may be easier to stick with the explicit equation already given for y. So you attain the normal vector

[itex] n = \frac{(2x,-1,2z)}{\sqrt{4x^2 + 4z^2 + 1}} [/itex]

Projecting on to the xz-plane we get;

[itex] |n.j| = \frac{1}{\sqrt{4x^2 + 4z^2 + 1}} [/itex]

So

[itex] dS = \frac{dA_y}{|n.j|} = \sqrt{4x^2 + 4z^2 + 1} dA_y [/itex]

After that it would be ideal to convert to polar co-ordinates. You also seem to have left the original y out of the surface integral.

I was intending to project onto xy plane becasue we can get z as a function of x and y... How would one know whether to project onto the standard xy or in your case the xz plane...

Can you explain this

Projecting on to the xz-plane we get;

[itex] |n.j| = \frac{1}{\sqrt{4x^2 + 4z^2 + 1}} [/itex]

 

[bugatti79]

OK, it makes sense to project onto the xz plane

I set up the integrand to be

[itex] \displaystyle \int_{0}^{2 \pi} \int_{0}^{2} r sin \theta \sqrt(4r^2+1) r dr d \theta [/itex]...?

 

[Silversonic]

OK, it makes sense to project onto the xz plane

I set up the integrand to be

[itex] \displaystyle \int_{0}^{2 \pi} \int_{0}^{2} r sin \theta \sqrt(4r^2+1) r dr d \theta [/itex]...?

Almost

[itex] y = x^2 + z^2 = r^2 [/itex]

So you get [itex] \displaystyle \int_{0}^{2 \pi} \int_{0}^{2} r^2 \times \sqrt(4r^2+1)dS [/itex]

Where dS is the surface of this circle, which is given by [itex] r dr d \theta [/itex]

[itex] \displaystyle \int_{0}^{2 \pi} \int_{0}^{2} r^2 \times \sqrt(4r^2+1) \times rdrd \theta [/itex]

You can use the substitution [itex] u = 4r^2 + 1 [/itex]

Wondered why this question seemed familiar, is it out of the James Stewart Calculus book?

 

[bugatti79]

Almost

[itex] y = x^2 + z^2 = r^2 [/itex]

So you get [itex] \displaystyle \int_{0}^{2 \pi} \int_{0}^{2} r^2 \times \sqrt(4r^2+1)dS [/itex]

Where dS is the surface of this circle, which is given by [itex] r dr d \theta [/itex]

[itex] \displaystyle \int_{0}^{2 \pi} \int_{0}^{2} r^2 \times \sqrt(4r^2+1) \times rdrd \theta [/itex]

You can use the substitution [itex] u = 4r^2 + 1 [/itex]

Wondered why this question seemed familiar, is it out of the James Stewart Calculus book?

Wolfram alpha seems to use u=r^2. Both give different answers.

My attempt for u=4^r2+1 is du=8rdr implies rdr =du/8 and r^2=(u-1)/4

The integrand becomes

[itex] \displaystyle \frac{1}{32} \int_0^2 u^{3/2} -u^{1/2} du[/itex]...?

 

[Silversonic]

Wolfram alpha seems to use u=r^2. Both give different answers.

My attempt for u=4^r2+1 is du=8rdr implies rdr =du/8 and r^2=(u-1)/4

The integrand becomes

[itex] \displaystyle \frac{1}{32} \int_0^2 u^{3/2} -u^{1/2} du[/itex]...?

Did you change your limits?

 

[bugatti79]

Did you change your limits?

Forgot that. Thanks