Find the surface area of the part of the hyperbolic paraboloid $z=x^{2}-y^{2}$ that lies within the cylinder $x^{2}+y^{2}=1$ in the first octant

View step-by-step help

Home

Math Problems

Algebra 2

Reflections of functions

Full solution

Q. Find the surface area of the part of the hyperbolic paraboloid

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

that lies within the cylinder

x^{2}+y^{2}=1

in the first octant

Identify Region: Identify the region of integration. $\newline$ Since we are dealing with the first octant and the cylinder $x^2 + y^2 = 1$ , the limits for $x$ and $y$ are both from $0$ to $1$ , and $x^2 + y^2 \leq 1$ .
Set Up Integral: Set up the double integral for the surface area. $\newline$ The surface area $S$ of $z = f(x, y)$ over a region $D$ is given by the integral $\int\int_D \sqrt{1 + (f_x)^2 + (f_y)^2} \, dA$ , where $f_x$ and $f_y$ are partial derivatives of $f$ with respect to $x$ and $y$ , respectively.
Calculate Derivatives: Calculate the partial derivatives. $\newline$ For $z = x^2 - y^2$ , $f_x = 2x$ and $f_y = -2y$ . Then, $(f_x)^2 = (2x)^2 = 4x^2$ and $(f_y)^2 = (-2y)^2 = 4y^2$ .
Substitute in Formula: Substitute into the surface area formula. $\newline$ The integrand becomes $\sqrt{1 + 4x^2 + 4y^2}$ . The integral for the surface area is $\int\int_D \sqrt{1 + 4x^2 + 4y^2} \, dA$ .
Convert to Polar: Convert to polar coordinates for easier integration. $\newline$ $x = r \cos(\theta)$ , $y = r \sin(\theta)$ , and $dA = r dr d\theta$ . The limits for $r$ are from $0$ to $1$ , and for $\theta$ from $0$ to $\pi/2$ since we are in the first octant.
Substitute Polar Coordinates: Substitute polar coordinates into the integral. The integral becomes $\int_{0}^{\pi/2} \int_{0}^{1} \sqrt{1 + 4r^2} r \, dr \, d\theta$ .
Integrate with Respect to $r$ : Perform the integration with respect to $r$ first. $\int_{0}^{1} r \sqrt{1 + 4r^2} \, dr$ . Let $u = 1 + 4r^2$ , then $du = 8r \, dr$ , and $r \, dr = \frac{du}{8}$ . The limits change to $u = 1$ when $r = 0$ and $u = 5$ when $r = 1$ .
Continue Integration: Continue the integration. $\newline$ The integral becomes $\frac{1}{8} \int_{1}^{5} \sqrt{u} \, du$ . This integral evaluates to $\frac{1}{8} \times \frac{2}{3} \times u^{\frac{3}{2}}$ from $1$ to $5$ .
Evaluate Integral: Evaluate the integral and multiply by the angular range. $\newline$ The integral of $u^{\frac{3}{2}}$ from $1$ to $5$ is $\frac{2}{3} \cdot (5^{\frac{3}{2}} - 1^{\frac{3}{2}})$ . Calculate $5^{\frac{3}{2}} = 11.1803$ and $1^{\frac{3}{2}} = 1$ . The result is $\frac{2}{3} \cdot (11.1803 - 1) = 6.7869$ . Then, $\frac{1}{8} \cdot 6.7869 = 0.8484$ . Multiply by the angular range $\frac{\pi}{2} = 1.5708$ . Final result is $$1$.$5708$ \cdot $0$.$8484$ = $1$.$332$.