Find the area of the surface. $\newline$ the part of the hyperbolic paraboloid $z=y^{2}-x^{2}$ that lies between the cylinders $x^{2}+y^{2}=4$ and $x^{2}+y^{2}=16$

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Find the axis of symmetry of a parabola

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Q. Find the area of the surface.

\newline

the part of the hyperbolic paraboloid

z=y^{2}-x^{2}

that lies between the cylinders

x^{2}+y^{2}=4

and

x^{2}+y^{2}=16

Surface Area Integral Formula: To find the surface area, we'll use the surface area integral for $z = f(x, y)$ . The formula is $\int\int\sqrt{1 + \left(\frac{dz}{dx}\right)^2 + \left(\frac{dz}{dy}\right)^2} dA$ , where $dA$ is the differential area element in the xy-plane.
Calculate Partial Derivatives: First, calculate the partial derivatives $\frac{dz}{dx}$ and $\frac{dz}{dy}$ . For $z = y^2 - x^2$ , $\frac{dz}{dx} = -2x$ and $\frac{dz}{dy} = 2y$ .
Plug Derivatives into Formula: Now, plug the derivatives into the formula: $\sqrt{1 + (-2x)^2 + (2y)^2} = \sqrt{1 + 4x^2 + 4y^2}$ .
Use Polar Coordinates: The region of integration is between the circles $x^2 + y^2 = 4$ and $x^2 + y^2 = 16$ . It's easier to use polar coordinates, where $x = r\cos(\theta)$ and $y = r\sin(\theta)$ .
Differential Area Element: In polar coordinates, the differential area element $dA$ is $r dr d\theta$ . The limits for $r$ are from $2$ to $4$ (the radii of the circles), and for $\theta$ from $0$ to $2\pi$ (full circle).
Substitute Polar Coordinates: Substitute $x$ and $y$ with polar coordinates in the integrand: $\sqrt{1 + 4r^2\cos^2(\theta) + 4r^2\sin^2(\theta)} = \sqrt{1 + 4r^2}$ .
Set Up Double Integral: Set up the double integral: $\int_{\theta=0}^{2\pi} \int_{r=2}^{4} r\sqrt{1 + 4r^2} \, dr \, d\theta$ .
Integrate with Respect to r: Integrate with respect to r first: $\int r\sqrt{1 + 4r^2} \, dr$ . Let $u = 1 + 4r^2$ , then $du = 8r \, dr$ . So, $\frac{1}{8} du = r \, dr$ .
Evaluate Integral for r: The integral becomes $\frac{1}{8} \int \sqrt{u} \, du$ , which is $\frac{1}{8} \cdot \frac{2}{3}u^{\frac{3}{2}}$ . Substitute back for u to get $\left(\frac{1}{12}\right)(1 + 4r^2)^{\frac{3}{2}}$ .
Integrate with Respect to $\theta$ : Evaluate the integral from $r=2$ to $r=4$ : $\frac{1}{12}(1 + 4(4)^2)^{\frac{3}{2}} - \frac{1}{12}(1 + 4(2)^2)^{\frac{3}{2}}$ .
Multiply by $2\pi$ : Calculate the values: $(\frac{1}{12})(1 + 64)^{\frac{3}{2}} - (\frac{1}{12})(1 + 16)^{\frac{3}{2}} = (\frac{1}{12})(65)^{\frac{3}{2}} - (\frac{1}{12})(17)^{\frac{3}{2}}$ .
Calculate Final Value: Now, integrate with respect to $\theta$ from $0$ to $2\pi$ . The integral is just $2\pi$ times the result from the $r$ integration because there's no $\theta$ in the integrand.
Simplify the Expression: Multiply by $2\pi$ : $2\pi \times \left[\left(\frac{1}{12}\right)(65)^{\frac{3}{2}} - \left(\frac{1}{12}\right)(17)^{\frac{3}{2}}\right]$ .
Plug in Values: Calculate the final value: $$2$\pi \times \left[\left(\frac{$1$}{$12$}\right)($65$)^{\frac{$3$}{$2$}} - \left(\frac{$1$}{$12$}\right)($17$)^{\frac{$3$}{$2$}}\right] = $2$\pi \times \left[\frac{($65$)^{\frac{$3$}{$2$}}}{$12$} - \frac{($17$)^{\frac{$3$}{$2$}}}{$12$}\right].
Final Calculation: Simplify the expression: $2\pi \times [(65)^{\frac{3}{2}} - (17)^{\frac{3}{2}}] / 12$.
Final Calculation: Simplify the expression: $2\pi \times [(65)^{\frac{3}{2}} - (17)^{\frac{3}{2}}] / 12$.Plug in the values and calculate the final answer: $2\pi \times [(65)^{\frac{3}{2}} - (17)^{\frac{3}{2}}] / 12 \approx 2\pi \times [520.9 - 70.1] / 12$.
Final Calculation: Simplify the expression: $2\pi \times [(65)^{\frac{3}{2}} - (17)^{\frac{3}{2}}] / 12$. Plug in the values and calculate the final answer: $2\pi \times [(65)^{\frac{3}{2}} - (17)^{\frac{3}{2}}] / 12 \approx 2\pi \times [520.9 - 70.1] / 12$. Final calculation: $2\pi \times 450.8 / 12 \approx 2\pi \times 37.57 \approx 236.2\pi$.