Question

Find the area of the surface. The portion of the cone \(z=2 \sqrt{x^{2}+y^{2}}\) inside the cylinder \(x^{2}+y^{2}=4\)

Short answer

The area of the surface defined by the portion of the cone \(z=2 \sqrt{x^{2}+y^{2}}\) inside the cylinder \(x^{2}+y^{2}=4\) is \(A = \frac{64}{3}\pi\).\n

Step-by-step solution

Identify the Parameters of the Surface

We first need to identify the surface in question. The surface is defined by the function \(z = 2 \sqrt{x^{2}+y^{2}}\) which represents a cone. We are looking specifically at the portion of the cone inside a cylinder, which is defined by \(x^{2} + y^{2} = 4\). This gives us a radius of 2 for our surface.

Parameterize the Surface

With our radius known, we can now parameterize our surface in terms of cylindrical coordinates (r, θ, z). Given \(x = r \cos(θ), y = r \sin(θ)\), we can substitute these into the equation for our cone to obtain: \(z = 2 r\). The boundaries of \(r\) and \(θ\) are determined by the cylinder: \(0 ≤ r ≤ 2\), and \(0 ≤ θ ≤ 2π\). Now, find the differential surface area in cylindrical coordinates, which is \(rdrdθ\).

Implement Surface Area Integral

The formula for the surface area of a parameterized surface like ours is given by the double integral \[A = \int \int \sqrt{1 + (f_{x})^{2} + (f_{y})^{2}} \, dS\] where \(f_{x}\) and \(f_{y}\) are the partial derivatives of \(z\) with respect to \(x\) and \(y\) respectively, and \(dS = rdrdθ\) is the differential surface area.

Compute the Surface Area Integral

To apply the formula, first calculate the partial derivatives, then compute the square root term, which with simplification will equal \(\sqrt{r^{2} + 4}\). Then integrate over \(r\) and \(θ\): \[A = \int_0^{2\pi} \int_0^2 r \sqrt{r^{2} + 4} \, dr\,dθ.\] This integral can be solved iteratively using standard integration techniques, such as u-substitution.

Final Calculation

The integral over r returns 32/3, which can then be multiplied by the integral over θ, resulting in the final surface area of the portion of the cone inside the cylinder as \(A = \frac{64}{3}\pi\).