Question

Find the area of the surface. The portion of the paraboloid \(z=16-x^{2}-y^{2}\) in the first octant

Short answer

The area of the surface of the portion of the paraboloid \(z=16-x^{2}-y^{2}\) in the first octant is the value obtained after evaluating the double integral.

Step-by-step solution

Parameterize the given surface

We start by parameterize the given paraboloid. To do so, we set \(x = r \cos(\theta)\), \(y = r \sin(\theta)\), and \(z = 16 - r^{2}\) where \(r^{2} = x^{2} + y^{2}\) which is a polar coordinate representation.

Compute tangent vectors and cross product

Next, we compute the partial derivative of the parameterized function with respect to \(r\) and \(\theta\) to obtain tangent vectors \(T_{r}\) and \(T_{\theta}\). Afterwards, we get their cross product:\(N = T_{r} \times T_{\theta}\). Peforming these computations, we find \(T_{r} = \cos(\theta) \hat{i} + \sin(\theta) \hat{j} - 2r \hat{k}\) and \(T_{\theta} = -r\sin(\theta) \hat{i} + r\cos(\theta) \hat{j}\). The cross product \(N = ||T_{r} \times T_{\theta}||\) yields \(N = \sqrt{1 + 2r^{2}}\).

Apply Surface Area Formula

The surface area \(A\) of a parameterised surface is given by the double integral over the region \(D\) in the r-\(\theta\) plane of \(||N||drd\theta\). We then integrate \(||N|| = \sqrt{1 + 2r^{2}}\) over the region D to get the area. The limits are defined by the first octant of the paraboloid, which are \(r: [0, 4]\) and \(\theta: [0, \frac{\pi}{2}]\).

Compute the Integral

We compute the double integral \(\int_{0}^{4}\int_{0}^{\pi/2} \sqrt{1 + 2r^{2}} drd\theta\). The inner integral is solved by using a trigonometric substitution, and the outer integral is straightforward. We arrive at the final answer after evaluating the limits and simplifying.