Question

Find the area of the given surface over the region \(R\). \(f(x, y)=-2 x+4 y^{2}+7\) over \(R,\) the triangle bounded by \(y=-x, y=x, 0 \leq y \leq 1\).

Short answer

The area of the surface over the region is \(\frac{2}{195} [(70)^{3/2} - (5)^{3/2}]\).

Step-by-step solution

Identify the Region R

The region R is defined as the triangle bounded by the lines \(y = -x\), \(y = x\), and \(y = 1\). Within these boundaries and given the constraint \(0 \leq y \leq 1\), the triangular region is defined within the limits \(-y \leq x \leq y\). The region's vertices are at \((0,0)\), \((1,1)\), and \((-1,1)\).

Set Up the Surface Area Integral

The formula for the surface area of a function \(f(x, y)\) over a region \(R\) is given by \[ \iint_R \sqrt{\left(\frac{\partial f}{\partial x}\right)^2 + \left(\frac{\partial f}{\partial y}\right)^2 + 1} \, dx \, dy. \]

Compute Partial Derivatives

For the function \(f(x,y) = -2x + 4y^2 + 7\), compute the partial derivatives: \( \frac{\partial f}{\partial x} = -2 \) and \( \frac{\partial f}{\partial y} = 8y \).

Evaluate the Integrand

Substitute the partial derivatives into the integrand formula: \[ \sqrt{(-2)^2 + (8y)^2 + 1} = \sqrt{4 + 64y^2 + 1} = \sqrt{65y^2 + 5}. \]

Set Up the Integral Limits and Evaluate

The limits for integration concerning \(y\) are from 0 to 1. For each fixed \(y\), the limits for \(x\) are from \(-y\) to \(y\). Thus, the integral setup is \[ \int_{0}^{1} \int_{-y}^{y} \sqrt{65y^2 + 5} \, dx \, dy. \]

Integrate Over x

Evaluate the inner integral: \[ \int_{-y}^{y} \sqrt{65y^2 + 5} \, dx = 2y\sqrt{65y^2 + 5}. \]

Integrate Over y

Evaluate the outer integral: \[ \int_{0}^{1} 2y \sqrt{65y^2 + 5} \, dy. \] Use a substitution \(u = 65y^2 + 5\), \(du = 130y \, dy\), \(y \cdot dy = \frac{1}{130} \, du\). This becomes \[ \int_{5}^{70} \frac{2}{130} \sqrt{u} \, du = \frac{1}{65} \cdot \frac{2}{3} u^{3/2} \bigg|_{5}^{70}. \]

Evaluate the Substitution Results

Calculate \[ \frac{1}{65} \cdot \frac{2}{3} [(70)^{3/2} - (5)^{3/2}]. \] This yields the final numeric result after evaluating the specific expressions.

Key concepts

Partial Derivatives

When dealing with multivariable functions, partial derivatives are essential to understanding how changes in one variable affect the function while keeping others constant. In this exercise, the function is \( f(x, y) = -2x + 4y^2 + 7 \). To find its partial derivatives, you differentiate separately with respect to \(x\) and \(y\):

• The partial derivative with respect to \(x\) \( \left( \frac{\partial f}{\partial x} \right) \) considers how \(f\) changes as \(x\) changes. Here, it is \(-2\).
• The partial derivative with respect to \(y\) \( \left( \frac{\partial f}{\partial y} \right) \) reveals how \(f\) changes as \(y\) changes, resulting in \(8y\).

Understanding these derivatives helps in computing the surface area integral, which is the next step in solving such problems.

Surface Area Integral

For a smooth surface defined by a function \( f(x, y) \) over a region \(R\), the surface area is calculated using a specific integral. This is given by:\[\iint_R \sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1} \, dx \, dy.\] This expression accounts for the contributions of the slope in both directions (\(x\) and \(y\)). When using this formula, each component under the square root addresses how the surface stretches wrt its base. It combines the effects of changes in both directions to provide an accurate surface area measurement.

Double Integral

Double integration is the process of calculating the integral over a two-dimensional area. In our exercise, we perform a double integral to find the surface area of the function over a triangular region. We tackle the problem by

• First integrating with respect to \(x\) over the bounds \(-y\) to \(y\).
• This results in a new expression dependent only on \(y\).
• Then integrating with respect to \(y\) over the bounds \(0\) to \(1\).

This iterative process considers the entire region \(R\) to calculate the desired surface area intricately.

Triangle Region R

The region \( R \) in this problem is a triangle, defined by its boundary equations: \(y = -x\), \(y = x\), and \(y = 1\). These lines intersect at specific points, forming a triangular area within which we evaluate our surface.

• The vertices of this triangle are at \((0,0)\), \((1,1)\), and \((-1,1)\).
• It is crucial to set proper limits for integration to correctly compute the integrals over this region.
• The constraints ensure that the calculations only occur within \(-y \leq x \leq y\) for \(0 \leq y \leq 1\), effectively bounding the integration.

This understanding is necessary for correctly setting up and solving the surface area integral over \( R \).