Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

The paraboloid \(z=x^{2}+y^{2}\) over the region (a) in the first quadrant and inside the circle \(x^{2}+y^{2}=9\) (b) inside the triangle with vertices \((0,0),(3,0),(0,3)\)

Step-by-Step Solution

Answer

(a) \(\frac{81\pi}{8}\), (b) \(\frac{27}{4}\)

Step 1: Understand the Problem

The exercise involves evaluating the paraboloid defined by the equation \( z = x^2 + y^2 \) over two specified regions in the plane.

Step 1: Define the Region for Part (a)

For part (a), the region is the first quadrant portion inside the circle described by \( x^2 + y^2 = 9 \). The circle has a radius of 3. So we consider the region where \( x, y \geq 0 \) and \( x^2 + y^2 \leq 9 \).

Step 2: Set Up the Double Integral for Part (a)

Convert to polar coordinates, where \( x = r\cos\theta \) and \( y = r\sin\theta \). The double integral becomes \( \int_{0}^{\frac{\pi}{2}} \int_{0}^{3} r^2 \cdot r \, dr \, d\theta \), which is \( \int_{0}^{\frac{\pi}{2}} \int_{0}^{3} r^3 \, dr \, d\theta \).

Step 3: Integrate with Respect to r for Part (a)

Compute the inner integral: \( \int_{0}^{3} r^3 \, dr = \left. \frac{r^4}{4} \right|_0^3 = \frac{81}{4} \).

Step 4: Integrate with Respect to θ for Part (a)

Now, compute the outer integral: \( \int_{0}^{\frac{\pi}{2}} \frac{81}{4} \, d\theta = \frac{81}{4} \cdot \left. \theta \right|_0^{\frac{\pi}{2}} = \frac{81\pi}{8} \).

Step 5: Define the Region for Part (b)

For part (b), the region is a triangular area with vertices at \((0,0), (3,0), (0,3)\), bounded by the lines \(y=0\), \(x=0\), and \(y=3-x\).

Step 6: Set Up the Double Integral for Part (b)

In Cartesian coordinates, the double integral is \( \int_{0}^{3} \int_{0}^{3-x} (x^2 + y^2) \, dy \, dx \).

Step 7: Integrate with Respect to y for Part (b)

Compute the inner integral: \( \int_{0}^{3-x} (x^2 + y^2) \, dy = \int_{0}^{3-x} x^2 \, dy + \int_{0}^{3-x} y^2 \, dy \).

Step 8: Solve the Inner Integrals for Part (b)

Calculate: \( \int_{0}^{3-x} x^2 \, dy = x^2(3-x) \) and \( \int_{0}^{3-x} y^2 \, dy = \left. \frac{y^3}{3} \right|_0^{3-x} = \frac{(3-x)^3}{3} \). Sum these results.

Step 9: Simplify the Inner Integral for Part (b)

Combine: \( x^2(3-x) + \frac{(3-x)^3}{3} \). This simplifies to \( 9x - 4x^2 + \frac{(3-x)^3}{3} \).

Step 10: Integrate with Respect to x for Part (b)

Compute the outer integral: \( \int_{0}^{3} \left( 9x - 4x^2 + \frac{(3-x)^3}{3} \right) \, dx \). Separate this into simpler integrals and solve.

Step 11: Calculate the Final Integral for Part (b)

Calculate: \( \int_{0}^{3} 9x \, dx = \left. \frac{9x^2}{2} \right|_0^3 = \frac{81}{2} \), \( \int_{0}^{3} -4x^2 \, dx = \left. -\frac{4x^3}{3} \right|_0^3 = -36 \), \( \int_{0}^{3} \frac{(3-x)^3}{3} \, dx = \frac{1}{3} \cdot \left(\frac{(3-x)^4}{4}\right|_0^3 = \frac{27}{4} \).

Step 12: Sum the Results for Part (b)

Combine the results: \( \frac{81}{2} - 36 + \frac{27}{4} \), giving a final result of \( \frac{81}{2} - \frac{144}{4} + \frac{27}{4} = \frac{27}{4} \).

Step 13: Conclude the Results

The integral value for part (a) is \( \frac{81\pi}{8} \) and for part (b) is \( \frac{27}{4} \).

Key Concepts

Paraboloid

A paraboloid is a three-dimensional surface that resembles the shape of a parabola, but extended into an additional dimension. In this context, we focus on the equation \( z = x^2 + y^2 \), which describes a circular paraboloid. This shape opens upwards and has rotational symmetry about the z-axis.

Visualize the paraboloid as the set of points in three-dimensional space where each z-coordinate equals the sum of the squares of its corresponding x and y coordinates. This property makes it simple to identify points on the surface:

• If \((x, y) = (0, 0)\), then \(z = 0\).
• If \((x, y) = (1, 1)\), then \(z = 1 + 1 = 2\).
• Similarly, if \((x, y) = (3, 0)\) or \((0, 3)\), both give \(z = 9\).

The values of x and y determine the 'height' of z above or below the xy-plane. This relationship forms an important foundation for understanding volumes underneath or around the paraboloid.

Polar Coordinates

Polar coordinates offer a convenient way to describe regions and solve problems involving circles or radial symmetry. Instead of using x and y, polar coordinates use r (the radial distance from the origin) and \(\theta\) (the angle from the positive x-axis).

Mathematically, the conversion to polar coordinates is given by:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)

In the context of double integrals, the differential area element \(dx \, dy\) transforms to \(r \, dr \, d\theta\). This transformation is pivotal when evaluating integrals over circular regions, as it simplifies the computation significantly.

For region (a), using polar coordinates simplifies the setup and integration process, as the region is described naturally as a sector of a circle. Here, \(r\) varies from 0 to 3, and \(\theta\) spans from 0 to \(\frac{\pi}{2}\). This setup reflects the first quadrant of the circular region with radius 3.

Cartesian Coordinates

In Cartesian coordinates, points are described using perpendicular axes, typically labeled \(x\) and \(y\). These coordinates are straightforward for representing rectangular or triangular regions and are often convenient for integration over linear boundaries.

For part (b) of the exercise, Cartesian coordinates are utilized to describe the triangular region bounded by the points \((0, 0)\), \((3, 0)\), and \((0, 3)\).

The integration limits are established by the lines that form the triangle's boundary:

• The base of the triangle along the x-axis gives \(y = 0\).
• The left side of the triangle along the y-axis gives \(x = 0\).
• The hypotenuse is described by the equation \(y = 3 - x\).

These constraints lead to setting up the double integral first over \(y\), from 0 to \(3-x\), and then over \(x\), from 0 to 3. This order reflects the natural boundaries determined by the triangle's edges.

Region of Integration

The region of integration is the specific area over which we evaluate a double integral. Identifying this region correctly is crucial for setting up and solving the integral effectively.

In problem (a), the region of integration is a quarter circle in the first quadrant, with a radius of 3. This means we integrate over a circular area cut by the axes, making it perfect for polar coordinates.

For problem (b), the triangular region bounded by points \((0,0)\), \((3,0)\), and \((0,3)\) requires a Cartesian approach, because of its linear boundaries. Each problem's region of integration determines the most fitting coordinate system to simplify calculations.

• Circular regions are often best handled with polar coordinates.
• Rectangular or triangular regions are more efficiently described using Cartesian coordinates.

Choosing the right region and coordinate system allows for a more straightforward integration process, reducing the complexity of the computations.