Problem 23 - Thomas Calculus

Question

Sketch the region of integration and evaluate the integral. \begin{equation} \int_{0}^{1} \int_{0}^{y^{2}} 3 y^{3} e^{x y} d x d y \end{equation}

Step-by-Step Solution

Answer

The integral evaluates to \( e - 2 \).

Step 1: Identify the Boundaries

The integral is given as \( \int_{0}^{1} \int_{0}^{y^{2}} 3 y^{3} e^{x y} \, d x \, d y \). The outer integral indicates that \( y \) ranges from 0 to 1, and for each fixed \( y \), the inner integral tells us that \( x \) ranges from 0 to \( y^{2} \). This means the region of integration is bounded by \( y = 0 \), \( y = 1 \), \( x = 0 \), and \( x = y^2 \).

Step 2: Sketch the Region of Integration

Sketch the region defined by the conditions of the integrals. It is bounded below by the \(x\)-axis, on the right by the line \(x = y^2\), and vertically from \(y = 0\) to \(y = 1\). The curve \(x = y^2\) is a parabola opening to the right, starting at the origin (0,0) and ending at (1,1) within the specified bounds.

Step 3: Inner Integral Evaluation

Evaluate the inner integral \( \int_{0}^{y^{2}} 3 y^{3} e^{x y} \, d x \). Treat \( y \) as a constant during this integration. The integral of \( e^{x y} \) with respect to \( x \) is \( \frac{e^{x y}}{y} \). Thus, the result is: \[ 3 y^{3} \left[ \frac{e^{x y}}{y} \right]_{0}^{y^2} = 3 y^{2} (e^{y^3} - 1). \]

Step 4: Outer Integral Evaluation

Now, substitute the result from the inner integral back into the outer integral: \[ \int_{0}^{1} 3 y^{2} (e^{y^3} - 1) \, d y. \] Break it into two integrals: \[ 3 \int_{0}^{1} y^{2} e^{y^3} \, d y - 3 \int_{0}^{1} y^{2} \, d y. \]

Step 5: Evaluate \( \int_{0}^{1} y^{2} e^{y^3} \, d y \)

Use substitution \( u = y^3 \), \( du = 3y^2 \, dy \). Hence, \( \frac{du}{3} = y^2 \, dy \). The limits change from \( y = 0 \) to \( u = 0 \), and from \( y = 1 \) to \( u = 1 \). This integral becomes: \[ \frac{1}{3} \int_{0}^{1} e^{u} \, du = \frac{1}{3} [e^{u}]_{0}^{1} = \frac{e - 1}{3}. \]

Step 6: Evaluate \( 3 \int_{0}^{1} y^{2} \, d y \)

Calculate the second integral directly: \[ 3 \int_{0}^{1} y^{2} \, d y = 3 \left[ \frac{y^3}{3} \right]_{0}^{1} = [y^3]_{0}^{1} = 1. \]

Step 7: Combine Results

Combine results from the previous steps: \[ 3 \left( \frac{e - 1}{3} \right) - 1 = e - 2. \]

Step 8: Final Answer

The evaluated integral \( \int_{0}^{1} \int_{0}^{y^{2}} 3 y^{3} e^{x y} \, d x \, d y \) gives the final result of \( e - 2 \).

Key Concepts

Region of Integration

When solving a double integral, one of the first steps is understanding the region over which you are integrating. This region is specified by the limits of integration. In this problem, the integral is written as \( \int_{0}^{1} \int_{0}^{y^{2}} 3 y^{3} e^{x y} \, d x \, d y \). The outer integral determines that \( y \) varies from 0 to 1, which defines the vertical boundaries. The inner integral specifies that for each value of \( y \), \( x \) ranges from 0 to \( y^2 \). These boundaries describe a region in the xy-plane, bounded by:

• The x-axis (where \( y = 0 \))
• The line \( y = 1 \)
• The line \( x = 0 \)
• The parabola \( x = y^2 \)

This creates a region under the curve \( x = y^2 \) and above the x-axis, limiting the horizontal span from 0 to each corresponding \( y^2 \) value, forming a curvilinear shape between the lines.

Parabola

A parabola is a symmetric curve formed by the points that are equidistant from a fixed point (focus) and a fixed straight line (directrix). In the context of this double integral, the parabola \( x = y^2 \) plays a crucial role in defining the boundaries of integration.
This specific parabola is one that opens to the right, as opposed to the more familiar up-and-down opening parabolas in the standard form \( y = x^2 \).

• The vertex of this parabola is at the origin (0,0).
• It intersects the line \( y=1 \) at the point (1,1).
• As \( y \) increases from 0 to 1, the corresponding \( x \) values trace a curve from (0,0) to (1,1).

Understanding the shape and orientation of the parabola helps in visualizing and correctly sketching the region over which the integration will take place.

Substitution Method

The substitution method is an integral technique that involves changing the variable of integration to simplify the integral's computation. In this exercise, after evaluating the inner integral, the substitution method is used to simplify the outer integral:
For the outer integral \( \int_{0}^{1} y^{2} e^{y^3} \, d y \), we introduce the substitution \( u = y^3 \). This implies that \( du = 3y^2 \, dy \), and thus \( y^2 \, dy = \frac{du}{3} \). After substitution, the integral becomes:

• \( \frac{1}{3} \int_{0}^{1} e^{u} \, du \)
• The limits of integration change with \( y = 0 \rightarrow u = 0 \) and \( y = 1 \rightarrow u = 1 \).

This simplification using substitution allows for the easy integration of the exponential function, resulting in \( \frac{e-1}{3} \), an essential intermediate step in solving the problem.

Definite Integral

A definite integral represents the area under a curve within specified bounds. For this problem, we began with a double integral, which essentially computes the volume under a surface over a given region. In stepwise fashion, we evaluate this by first finding the inner integral within its bounds, followed by the outer integral.
Once both integrals are evaluated, as shown in the original solution, the final definite integral provides us with a numerical value that represents the computed volume. In this case, the outcome of combining both integrals, is \( e - 2 \).
The power of definite integrals is in their ability to translate complex shapes and regions into single, understandable numeric values, highlighting the elegance and utility of this fundamental calculus tool in determining area, volume, and other integral based quantities.