Problem 57 - Thomas Calculus

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 of the region is \(\frac{8}{3}\).

Step 1: Visualize the region

The paraboloid is given by the equation \(z = x^2 + y^2\), which opens upwards from the origin. The region in the \(xy\)-plane is a triangle defined by the lines \(y = x\), \(x = 0\), and \(x + y = 2\). Sketch this triangular region in the first quadrant.

Step 2: Define the Region of Integration

The triangle is in the first quadrant and its vertices are at \((0,0)\), \((2,0)\), and \((0,2)\). The bounds for \(x\) are from \(0\) to \(2\). For a fixed \(x\), \(y\) ranges from \(x\) (the line \(y=x\)) to \(2-x\) (the line \(x+y=2\)).

Step 3: Setup the Double Integral

The volume under the surface \(z = x^2 + y^2\) above the triangular region is calculated using a double integral. Thus, \[ V = \int_{0}^{2} \int_{x}^{2-x} (x^2 + y^2) \, dy \, dx. \]

Step 4: Integrate with respect to y

Integrate the function \(x^2 + y^2\) with respect to \(y\) from \(y = x\) to \(y = 2-x\). This yields:\[ \int_x^{2-x} (x^2 + y^2) \, dy = [x^2y + \frac{y^3}{3}]_x^{2-x}. \]Evaluate at the bounds to get:\[ = x^2(2-x) + \frac{(2-x)^3}{3} - \left[x^2 \cdot x + \frac{x^3}{3}\right]. \]

Step 5: Simplify the Result

Simplify the expression from the previous step:\[ = x^2(2-x) + \frac{(2-x)^3}{3} - (x^3 + \frac{x^3}{3}) \]\[ = 2x^2 - x^3 + \frac{8 - 12x + 6x^2 - x^3}{3} - \frac{4x^3}{3}. \]Combine like terms and simplify.

Step 6: Integrate with respect to x

Now integrate the simplified expression with respect to \(x\) from \(0\) to \(2\):\[ V = \int_{0}^{2} \left( 2x^2 - x^3 + \frac{8 - 12x + 6x^2 - x^3}{3} \right) \, dx. \] Expand and simplify the terms where necessary and then integrate term by term.

Step 7: Compute the Integral

Evaluate the definite integral from the previous step. Substitute the upper limit \(2\) and lower limit \(0\), then compute:\[ \left[ \frac{2x^3}{3} - \frac{x^4}{4} + \frac{8x}{3} - 2x^2 + \frac{2x^3}{3} - \frac{x^4}{3} \right]_0^2. \] Calculate each term and simplify to find the total volume.

Key Concepts

Paraboloid

A **paraboloid** is a three-dimensional surface that has the shape of a parabola revolved around its axis. In this particular exercise, the paraboloid is described by the equation \( z = x^2 + y^2 \). This equation represents an "elliptic paraboloid" because it opens upwards, widening from the origin as both \( x \) and \( y \) increase.
Key characteristics of a paraboloid include:

• It is symmetrical about its vertical axis, which in this case is the z-axis.
• As \( x \) and \( y \) move away from the origin, the values of \( z \) increase, indicating the surface rises.
• The intersection of the paraboloid with higher horizontal planes results in circular cross-sections.

This shape is essential for the problem as it defines the upper boundary of the volume we wish to calculate.

Double Integration

**Double integration** is a method used to calculate the volume under a surface over a two-dimensional region. In this exercise, double integration helps find the volume defined by the paraboloid above and the given triangular region below.
The double integration set up involves two integrals:

• The inner integral, \( \int (x^2 + y^2) \, dy \), calculates the area under the curve for a fixed \( x \).
• The outer integral, \( \int \, dx \), sums these areas as \( x \) varies across the defined region.

The integration bounds for \( y \) depend on the value of \( x \), which is a characteristic feature of double integrals when used over complex regions like triangles.

Region of Integration

The **region of integration** is essential to correctly set up the integrals and ensure the volume calculation reflects the actual bounded area. In our exercise, the region is the triangular section in the \( xy \)-plane, enclosed by the lines \( y = x \), \( x = 0 \), and \( x + y = 2 \).
This triangle has vertices at:

• \( (0, 0) \), where \( y = x \) meets \( x = 0 \)
• \( (2, 0) \), where \( x + y = 2 \) meets the \( x \)-axis
• \( (0, 2) \), where \( x + y = 2 \) meets the \( y \)-axis

Inside this triangle, for any given \( x \), \( y \) ranges from the line \( y = x \) to \( y = 2-x \). This ensures the integration captures the area precisely under the surface up to the limits laid out by these lines.

Definite Integrals

**Definite integrals** are used to calculate finite quantities, like the area or volume of a region. They represent the accumulation of quantities, such as summing infinitesimal slices of area or volume.
In this problem, two definite integrals are utilized:

• The first (inner) integral integrates over \( y \), calculating a slice of volume for a particular value of \( x \).
• The second (outer) integral sums these slices along the \( x \)-axis, from \( x = 0 \) to \( x = 2 \).

By evaluating the definite integrals, you get exact amounts since you are computing over a specific region with strict boundaries. The result gives us the exact volume of the region bounded by our paraboloid and triangular base.