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ishanz
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Homework Statement
Evaluate the surface integral.
∫∫S x^2*z^2 dS
S is the part of the cone z^2 = x^2 + y^2 that lies between the planes z = 1 and z = 3.
Homework Equations
[itex]\int \int _{S}F dS = \int \int _D F(r(u,v))|r_u\times r_v|dA[/itex]
[itex]x=rcos(\theta)[/itex]
[itex]y=rsin(\theta)[/itex]
The Attempt at a Solution
First, I parametrized the cone.
[itex]z^2=x^2+y^2\Rightarrow z=\sqrt{x^2+y^2} \Rightarrow z=r[/itex]
Therefore, the cone's vector equation should be
[itex]{\bf R}(r,\theta)=rcos(\theta){\bf i}+rsin(\theta){\bf j}+r{\bf k}[/itex]
[itex]{\bf R}_r=cos(\theta){\bf i}+sin(\theta){\bf j}+{\bf k}[/itex]
[itex]{\bf R}_\theta=-rsin(\theta){\bf i}+rcos(\theta){\bf j}[/itex]
[itex]|{\bf R}_r \times {\bf R}_\theta| = r\sqrt{2}[/itex]
[itex]\int_0^{2\pi}\int_1 ^3 {(rcos(\theta))^2(r)^2(r\sqrt{2})}drd\theta=\frac{364\sqrt{2}\pi}{3}[/itex]
Now, this is the right answer as per the book. My question is, when we go from [itex]dA[/itex] to [itex]drd\theta[/itex], why don't we use the Jacobian for polar coordinates, r? Had we included the Jacobian, the degree of r in the final double integral would have been six instead of five, giving us a completely different answer. My confusion here: Why is [itex]dA[/itex] not equal to [itex]rdrd\theta[/itex]?
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