Surface Integral of F on Cone S 0≤z≤4

In summary, the problem involves finding a surface integral for the cone described by z=\sqrt{x^2+y^2} with 0\leq z \leq 4. The surface integral is given by \int\int_S \textbf{F} \bullet d\textbf{S}, where the orientation of the surface S is given by the inner normal vector. To solve this, the surface S is parameterized as \textbf{r}(u,v)=u\textbf{i}+v\textbf{j}+\sqrt{u^2+v^2}\textbf{k} and the cross product \textbf{r}_u \times \textbf{r}_v is calculated.
  • #1
kudoushinichi88
129
2
I'm working on this problem,

Let S be the cone described by

[tex]z=\sqrt{x^2+y^2}[/tex]

where [itex]0\leq z \leq 4[/itex]

If

[tex]\textbf{F}(x,y,z)=y\textbf{i}-x\textbf{j}+z^2\textbf{k}[/tex]

find the surface integral

[tex]\int\int_S \textbf{F} \bullet d\textbf{S}[/tex]

where the orientation of S is given by the inner normal vector.So I first parameterized S into

[tex]\textbf{r}(u,v)=u\textbf{i}+v\textbf{j}+\sqrt{u^2+v^2}\textbf{k}[/tex]

Then I find the cross product

[tex]\textbf{r}_u \times \textbf{r}_v = -\frac{u}{\sqrt{u^2+v^2}}\textbf{i}-\frac{v}{\sqrt{u^2+v^2}}\textbf{j}+\textbf{k}[/tex]

So when I plug this into the surface integral formula, do I take the negative of this cross product? Because the orientation of S is given by the inner normal vector.
 
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  • #2
The answer should be 48\pi.The surface integral is \int_0^4 \int_0^{2\pi} \textbf{F} \bullet (-\frac{u}{\sqrt{u^2+v^2}}\textbf{i}-\frac{v}{\sqrt{u^2+v^2}}\textbf{j}+\textbf{k}) \rho \ d\rho \ d\thetawhere \rho=\sqrt{u^2+v^2}I just want to make sure I'm doing this correctly.
 

1. What is a surface integral?

A surface integral is a type of integration used to calculate the area of a surface in three-dimensional space. It involves integrating a function over a surface, using a double integral.

2. What is a cone?

A cone is a three-dimensional geometric shape with a circular base and a curved surface that tapers to a point at the top. It can be thought of as a stack of circles that get smaller and smaller towards the top.

3. How is the surface integral of F on a cone calculated?

The surface integral of F on a cone is calculated by integrating the function F over the surface of the cone. This involves setting up a double integral using the cone's parameters and the function's variables.

4. What does the constraint 0≤z≤4 mean in relation to the cone?

The constraint 0≤z≤4 means that the surface integral is only being calculated on the portion of the cone that lies between the z-values of 0 and 4. This can also be thought of as calculating the surface integral on the cone's height of 4 units.

5. What is the significance of calculating the surface integral of F on a cone?

Calculating the surface integral of F on a cone can provide valuable information about the cone's surface area, which can be useful in many fields such as physics, engineering, and geometry. It can also be used to solve problems involving the flow of fluids or heat on a conical surface.

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