Problem 10 - Calculus

Question

Find \(N\) and \(d S=|\mathbf{N}| d x d y\) and the surface area \(\iint d S\) Integrate over the \(x y\) shadow which ends where the \(z\) 's are equal \(\left(x^{2}+y^{2}=4\right.\) in Problem 1). \(z=x+y\) above triangle with vertices (0,0),(2,2),(0,2).

Step-by-Step Solution

Answer

The surface area is \(2\sqrt{3}\).

Step 1: Define the Surface Equation

We have been given the surface equation as \(z = x + y\). We need to integrate over the projection of this surface on the \(xy\)-plane.

Step 2: Find the Normal Vector

The vector normal to the surface \(z = f(x, y) = x + y\) can be found using the gradient: \(abla f = (-f_x, -f_y, 1)\). Here, \(f_x = 1\) and \(f_y = 1\), so \(\mathbf{N} = (-1, -1, 1)\).

Step 3: Compute Magnitude of the Normal Vector

The magnitude of the normal vector \(\mathbf{N} = (-1, -1, 1)\) is \(|\mathbf{N}| = \sqrt{(-1)^2 + (-1)^2 + 1^2} = \sqrt{3}\).

Step 4: Set Up Surface Integral

The element of surface area is \(dS = |\mathbf{N}| \cdot dx \cdot dy = \sqrt{3} \cdot dx \cdot dy\). The surface area \(A\) can be found using the integral \(\iint dS\).

Step 5: Determine Region of Integration

The projection of the surface on the \(xy\)-plane is a triangle with vertices (0,0), (2,2), and (0,2). This triangle lies within \(y = 0\) to \(y = 2\) and for each \(y\), \(x\) ranges from 0 to \(y\).

Step 6: Compute the Surface Area Integral

Integrate \(\sqrt{3} \cdot dx \cdot dy\) over the triangular region. The integral is \[ \int_{0}^{2} \int_{0}^{y} \sqrt{3} \; dx \; dy = \sqrt{3} \cdot \int_{0}^{2} \left[ x \right]_{x=0}^{x=y} dy = \sqrt{3} \cdot \int_{0}^{2} y \, dy \].

Step 7: Evaluate the Integral

Evaluate \[\sqrt{3} \cdot \int_{0}^{2} y \, dy = \sqrt{3} \cdot \left[ \frac{y^2}{2} \right]_{0}^{2} = \sqrt{3} \cdot (\frac{4}{2}) = \sqrt{3} \cdot 2 = 2 \sqrt{3}.\]

Key Concepts

Normal Vector Calculation

The normal vector is a key player in surface integration as it helps define the geometry of the surface. When faced with a surface given by a function, like \(z = x + y\), the normal vector \(\mathbf{N}\) can be determined using the gradient \(abla f\) of \(f(x, y) = x + y\). This leads to \(\mathbf{N} = (-f_x, -f_y, 1)\).
This step involves computing the partial derivatives, which are \(f_x = 1\) and \(f_y = 1\) in this example. Applying these, the normal vector becomes \(\mathbf{N} = (-1, -1, 1)\).
Understanding the components of the normal vector is crucial, as it defines the orientation of the surface in space, which impacts how we calculate the surface area and other properties.

Setting up Surface Integrals

Surface integrals allow you to calculate quantities over a surface, such as area or flux. We use the calculated normal vector's magnitude to set up the surface integral. Here, the magnitude is \(|\mathbf{N}| = \sqrt{3}\). The formula for the differential surface area element \(dS\) becomes \(\sqrt{3} \, dx \, dy\).
This step transforms a three-dimensional surface problem into a manageable two-dimensional integral. Understanding this connection bridges the geometry of surfaces with practical calculations.

Region of Integration

The region of integration is defined by projecting the surface onto the \(xy\)-plane. In this exercise, the region is a triangle with vertices at points (0,0), (2,2), and (0,2). Recognizing this triangular region helps in specifying the limits for the integral
The triangle is bounded by lines \(y = 0\), \(y = x\), and \(y = 2\). For every \(y\), \(x\) ranges from 0 up to \(y\), which forms the integration limits: \( \int_{0}^{2} \int_{0}^{y} \, \sqrt{3} \, dx \, dy\).

Surface Area Calculation

Calculating the surface area involves evaluating the integral over the specified region. In this exercise, we compute: \[ \int_{0}^{2} \int_{0}^{y} \sqrt{3} \, dx \, dy = \sqrt{3} \, \int_{0}^{2} \left[ x \right]_{x=0}^{x=y} \, dy = \sqrt{3} \, \int_{0}^{2} y \, dy \]
By integrating \(\sqrt{3}\) from 0 to \(y\), and subsequently integrating \(y\) from 0 to 2, we evaluate the surface area.

• The solution yields \(\sqrt{3} \, 2\), resulting in an area of \(2\sqrt{3}\).
• Surface area calculations like this are fundamental in various applications, from physics to engineering.