Chapter 16: Problem 40
Find the area of the portion of the surface \(x^{2}-2 z=0\) that lies above the triangle bounded by the lines \(x=\sqrt{3}, y=0,\) and \(y=x\) in the \(x y\) -plane.
Short Answer
Expert verified
The area is \(\frac{7}{3}\).
Step by step solution
01
Understand the Surface Equation
The surface equation is given as \(x^2 - 2z = 0\). This can be rewritten to express \(z\) in terms of \(x\): \(z = \frac{x^2}{2}\). This tells us the surface is a paraboloid opening upwards in the \(xz\)-plane.
02
Identify the Region in the xy-plane
The region in the \(xy\)-plane where we're interested in finding the area above is bounded by the lines \(x = \sqrt{3}\), \(y = 0\), and \(y = x\). This forms a right triangle with vertices at \((0,0)\), \((\sqrt{3}, 0)\), and \((\sqrt{3}, \sqrt{3})\).
03
Set Up the Area Integral
To find the area of the portion of the surface above the triangular region in the \(xy\)-plane, we set up a double integral to evaluate the area of the surface. The area function for a surface given by \( z = f(x,y) \) is calculated using the formula: \[ A = \iint_R \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dx \, dy \] where \(R\) is the region in the \(xy\)-plane.
04
Derive the Partial Derivatives
Find the partial derivatives of \(z\): \(z = \frac{x^2}{2}\). We compute \(\frac{\partial z}{\partial x} = x\) and \(\frac{\partial z}{\partial y} = 0\) because \(z\) does not depend on \(y\).
05
Compute the Integral
Substitute the partial derivatives into the area formula: \[ A = \iint_R \sqrt{1 + x^2} \, dx \, dy \]Evaluate this over the triangular region: from \(x = 0\) to \(\sqrt{3}\) and \(y = 0\) to \(y = x\). \[ A = \int_{0}^{\sqrt{3}} \int_{0}^{x} \sqrt{1 + x^2} \, dy \, dx \]Execute the \(dy\) integration first, which is straightforward since the integrand is constant with respect to \(y\):\[ A = \int_{0}^{\sqrt{3}} x \sqrt{1 + x^2} \, dx \]This can be solved using substitution or recognizing it as a form suitable for substitution \( u = 1 + x^2 \).
06
Solve the Integral
Perform the substitution \(u = 1 + x^2\), \(du = 2x dx\), hence \(x dx = \frac{1}{2} du \). The limits change as follows: when \(x = 0, u = 1\); when \(x = \sqrt{3}, u = 4\). Thus,\[ A = \frac{1}{2} \int_{1}^{4} \sqrt{u} \, du \]Evaluating, we get:\[ A = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{4} = \frac{1}{3} [u^{3/2}]_{1}^{4} = \frac{1}{3} [8 - 1] = \frac{7}{3} \]
07
Conclusion
The area of the portion of the surface \(x^2 - 2z = 0\) that lies above the triangle in the \(x y\)-plane is \(\frac{7}{3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Double Integral
When we talk about finding areas of surfaces over a specific region, double integrals come into play. A double integral allows us to compute an integral over a two-dimensional area, which in this case is a triangle.
- The double integral is denoted by \[\iint_R f(x, y) \, dx \, dy\]where \( R \) defines the region over which we want to integrate.
- In many surface area calculations, like the exercise provided, the double integral calculates the area by considering the contribution from every infinitesimally small piece of the region.
- Each piece is essentially a tiny rectangle, and the sum of all these rectangles approximates the total area.
Partial Derivatives
Partial derivatives are a fundamental concept in multivariable calculus. They help us understand how a function changes as one variable changes while others are held constant.For a function \( z = f(x, y) \), like our paraboloid \( z = \frac{x^2}{2} \), we can find partial derivatives to understand the surface slope in different directions:
- \( \frac{\partial z}{\partial x} \): This derivative considers changes in \( z \) while only \( x \) changes. In our exercise, it simplifies to \( x \).
- \( \frac{\partial z}{\partial y} \): This derivative considers changes in \( z \) while only \( y \) changes. Since \( z \) is independent of \( y \), it simplifies to \( 0 \).
Paraboloid
A paraboloid is a 3D shape formed by a parabola that revolves around an axis. In the context of this exercise, we are dealing with a specific type called an "elliptic paraboloid." The given surface equation \( x^2 - 2z = 0 \) or equivalently \( z = \frac{x^2}{2} \) represents a paraboloid opening upwards in the \( xz \)-plane. What does this mean? The surface curves upwards and widens as \( x \) moves away from zero.
- The axis of symmetry for this paraboloid lies along the \( z \)-axis, and cross-sections of the surface are parabolas.
- Every slice parallel to the \( x \)-axis looks like a parabola.
- In our exercise, the paraboloid only comes into play above the selected triangular region in the \( xy \)-plane.