Surface Area Using Double Integrals
I need to find the surface area of the function:
$$z = y^2-x^2$$
Between the cylinders:
$$x^2+y^2=1$$ $$x^2+y^2=4$$
So far I've tried converting it into polar coordinates to no avail and I don't know how to integrate: $$2\int_1^2\int_\sqrt{1-y^2}^\sqrt{4-y^2}\sqrt{4x^2+4y^2+1}\,\mathrm{d}x\mathrm{d}y$$
I'm using $S(A)=\int\int\sqrt{F_x^2+F_y^2+1}\,\mathrm{d}A$
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$\begingroup$Those bounds you wrote don't make sense. If the bounds for $y$ are $1 < y < 2$, then $1-y^2 < 0$ and so $\sqrt{1-y^2}$ (the lower bound for $x$) is undefined.
Using the formula $S(A) = \displaystyle\iint\sqrt{F_x^2+F_y^2+1}\,dA$, you got $\displaystyle\iint\sqrt{4x^2+4y^2+1}\,dx\,dy$.
Evaluating this integral is easiest if you convert it to polar coordinates first.
For the integrand, note that $x^2+y^2 = r^2$. So $\sqrt{4x^2+4y^2+1} = $?
Also, the Jacobian of the polar coordinate transformation is $r$, so $\,dx\,dy = r\,dr\,d\theta$.
For the bounds, draw a picture. The region in the $xy$-plane between the circles $x^2+y^2=1$ and $x^2+y^2=4$ is a doughnut. What is inner and outer radius of this doughnut? That will give you the bounds for $r$. The bounds for $\theta$ should be easy to get.
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