Problem 40 - Thomas Calculus

Question

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.

Step-by-Step Solution

Answer

The area is \(\frac{7}{3}\).

Step 1: 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.

Step 2: 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 \((\sqrt{3}, \sqrt{3})\).

Step 3: 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.

Step 4: 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 \(y\).

Step 5: 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 \( u = 1 + x^2 \).

Step 6: 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:\[ A = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{4} = \frac{1}{3} [u^{3/2}]_{1}^{4} = \frac{1}{3} [8 - 1] = \frac{7}{3} \]

Step 7: Conclusion

The area of the portion of the surface \(x^2 - 2z = 0\) that lies above the triangle in the \(x y\)-plane is \(\frac{7}{3}\).

Key Concepts

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.

To set it up geometrically, we can think of projecting the surface onto the base region we are concerned with, like shining a light directly down onto the surface. The double integral helps sum up the area contributions of all those tiny surface projections over the region.For the given exercise, the region \( R \) formed by the intersection of lines \( x = \sqrt{3} \), \( y = 0 \), and \( y = x \) is a right triangle. The double integral calculates the area of the surface above this triangle.

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 \( 0 \).

Understanding these derivatives allows us to see how the shape of the surface changes in each spatial direction. By applying these partial derivatives to the surface area integral, they contribute to forming a clearer picture of how the surface's "steepness" affects the area calculation.

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.

Visualizing the paraboloid is key to understanding the surface's interaction with the \( xy \)-plane and the volume it encloses. Through the concept of double integrals and partial derivatives, we see how to compute complex surface areas by understanding a straightforward surface like a paraboloid.