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mathmari
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Hey!
If $0\leq a<b$, I want to calculate the area of the conic surface that is defined by the relations $z^2=x^2+y^2$ and $a\leq z\leq b$ in two ways:
I have the following:
If $0\leq a<b$, I want to calculate the area of the conic surface that is defined by the relations $z^2=x^2+y^2$ and $a\leq z\leq b$ in two ways:
- using cartesian coordinates
- using cylindrical coordinates
I have the following:
- We have that $z^2=x^2+y^2\Rightarrow z=\pm \sqrt{x^2+y^2}$. Since $z\geq a\geq 0$ we have that $z=\sqrt{x^2+y^2}$.
We have that $$a\leq z\leq b\Rightarrow a\leq \sqrt{x^2+y^2}\leq b \Rightarrow a^2\leq x^2+y^2\leq b^2 \Rightarrow a^2-x^2\leq y^2\leq b^2-x^2 \Rightarrow \sqrt{a^2-x^2}\leq y\leq \sqrt{b^2-x^2}$$
I am not really sure if we can just take the root of the inequality. (Wondering)
If this is correct, then the square roots are defined if $a^2-x^2\geq 0$ and $b^2-x^2\geq 0$. Since $a<b$ we get that $-a\leq x\leq a$.
So, we define the function $\Sigma : D \rightarrow \mathbb{R}^3$ with $\Sigma (x,y)=(x,y, x^2+y^2)$, where $D=[-a,a]\times[\sqrt{a^2-x^2}, \sqrt{b^2-x^2}]$.
We have that $\Sigma_x=(1,0,2x)$ and $\Sigma_y=(0,1,2y)$.
So, we get $\Sigma_x\times\Sigma_y=(-2x, -2y, 1)$ and so $\|\Sigma_x\times\Sigma_y\|=\sqrt{4x^2+4y^2+1}$.
So, we get $$A(\Sigma (D))=\iint_D\|\Sigma_x\times\Sigma_y\|dxdy=\int_{-a}^a\int_{\sqrt{a^2-x^2}}^{\sqrt{b^2-x^2}}\sqrt{4x^2+4y^2+1}dydx$$ Is everything correct so far? If yes, how could we continue? (Wondering)