Problem 57 - University Calculus: Early Transcendentals

Question

Find the volume of the region bounded above by the paraboloid \(z=x^{2}+y^{2}\) and below by the triangle enclosed by the lines \(y=x, x=0,\) and \(x+y=2\) in the \(x y\) -plane.

Step-by-Step Solution

Answer

The volume is \(\frac{4}{3}\).

Step 1: Understand the 3D region

The region is between the paraboloid surface given by \(z = x^2 + y^2\) and the triangular base located in the \(xy\)-plane. The base triangle has vertices at (0,0), (2,0), and (0,2), formed by the lines \(y = x\), \(x = 0\), and \(x + y = 2\).

Step 2: Set up the double integral

We will set up a double integral to find the volume beneath the surface in the \(z\)-direction. The volume \(V\) can be given by \(V = \int \int_R (x^2 + y^2) \, dx \, dy\), where \(R\) is the region in the \(xy\)-plane bounded by \(y = x\), \(x = 0\), and \(x+y = 2\).

Step 3: Determine the limits of integration

The region \(R\) is a triangle on the \(xy\)-plane. For a given \(x\), \(y\) ranges from \(y = x\) to \(y = 2 - x\). \(x\) ranges from \(0\) to \(2\). Thus, the limits for \(x\) are 0 to 2, and for each \(x\), the limits for \(y\) are \(x\) to \(2 - x\).

Step 4: Compute the inner integral with respect to y

Calculate the inner integral, which is \(\int_x^{2-x} (x^2 + y^2)\, dy\). This results from integrating with respect to \(y\), \(\left[ yx^2 + \frac{y^3}{3} \right]_x^{2-x}\).

Step 5: Simplify the result of the inner integral

Substitute the limits \(y = 2-x\) and \(y = x\) into the integrated function and simplify: \((2-x)x^2 + \frac{(2-x)^3}{3} - \left( x^3 + \frac{x^3}{3} \right)\).

Step 6: Compute the outer integral with respect to x

Integrate the result from Step 5 with respect to \(x\) over the interval from 0 to 2. Calculate \(\int_0^2 \left[(2-x)x^2 + \frac{(2-x)^3}{3} - \left( x^3 + \frac{x^3}{3} \right)\right] \, dx\).

Step 7: Evaluate the definite integrals

Compute the definite integrals from Step 6 to find the volume. After calculating and simplifying, you get: \(V = \frac{4}{3}\).

Key Concepts

Double Integral

A double integral is a way to integrate over a two-dimensional region. It is used to find the total accumulated value over the area, like volume, area, or other physical quantities that depend on the dimensions of the region. In this exercise, we use a double integral to find the volume beneath a surface over a specified area in the plane.

In mathematical terms, the double integral is expressed as \(\int \int_R f(x, y) \, dx \, dy\), where \(R\) represents the region of integration, and \(f(x, y)\) is the function to be integrated.

• The inner integral is with respect to \(y\), holding \(x\) constant. It represents the integral along the \(y\)-axis for each fixed \(x\).
• The outer integral is with respect to \(x\). It sums up the outcomes of the inner integrals across changes in \(x\).

This approach lets you compute the volume of a three-dimensional region by integrating the function that describes its height across the plane in which it is based.

Paraboloid

A paraboloid is a three-dimensional surface described by an equation similar to a parabola. The particular paraboloid given by the equation \(z = x^2 + y^2\) opens upwards, forming a bowl-like shape.

• The surface is symmetrical around the \(z\)-axis, and every cross-section parallel to the \(xy\)-plane is a circle.
• The vertex, or the lowest point in this configuration, is at the origin (0, 0, 0).
• As \(x\) and \(y\) move away from the origin, \(z\) increases, creating the curved surface of the paraboloid.

In our problem, we are interested in the volume under this paraboloid, bounded by a specified region in the \(xy\)-plane. The function \(z=x^2+y^2\) gives the height of the paraboloid above each point \((x, y)\) along the defined region.

Limits of Integration

The limits of integration define the boundaries for evaluating the integral. For a double integral, these limits determine the region over which the function is integrated.

For our problem, the region \(R\) is a triangular area in the \(xy\)-plane bounded by the lines \(y = x\), \(x = 0\), and \(x + y = 2\). To set the limits correctly, it's crucial to:

• Determine the range of \(x\), which in this problem goes from 0 to 2, based on the intercepts of the lines forming the triangle.
• For each specific \(x\), find the corresponding range of \(y\). Here, \(y\) changes from \(y = x\) to \(y = 2 - x\), which are given by the boundary lines.

These limits ensure the integral covers the entire desired region accurately, capturing the entire volume beneath the paraboloid over the triangular region.