Problem 8 - Thomas Calculus

Question

In Exercises \(1-10,\) sketch the region of integration and evaluate the integral. $$ \int_{1}^{2} \int_{y}^{y^{2}} d x d y $$

Step-by-Step Solution

Answer

The area of the region of integration is \( \frac{11}{6} \).

Step 1: Understand the Region of Integration

The limits of integration are from 1 to 2 for \( y \) and from \( y \) to \( y^2 \) for \( x \). This implies that for each \( y \), the value of \( x \) ranges from \( y \) to \( y^2 \). We are dealing with a region bounded by the lines \( x = y \) and \( x = y^2 \).

Step 2: Sketch the Region

Plot the lines \( x = y \) and \( x = y^2 \) on the coordinate plane. The curve \( x = y^2 \) is a parabola opening to the right, and \( x = y \) is a straight line. The region of integration is bounded below by these curves from \( y = 1 \) to \( y = 2 \).

Step 3: Set Up the Integral

The double integral is given as \( \int_{1}^{2} \int_{y}^{y^{2}} 1 \, dx \, dy \), where the integrand is simply 1. This represents the area of the region.

Step 4: Evaluate the Inner Integral

For the inner integral \( \int_{y}^{y^2} 1 \, dx \), integrate 1 with respect to \( x \): \[ \int_{y}^{y^2} 1 \, dx = [x]_{y}^{y^2} = y^2 - y.\]

Step 5: Evaluate the Outer Integral

Substitute the result of the inner integral into the outer one:\[ \int_{1}^{2} (y^2 - y) \, dy.\] Evaluate this integral:\[ = \left[ \frac{y^3}{3} - \frac{y^2}{2} \right]_{1}^{2} = \left( \frac{8}{3} - 2 \right) - \left( \frac{1}{3} - \frac{1}{2} \right).\] Simplify:\[ = \frac{8}{3} - 2 - \frac{1}{3} + \frac{1}{2} = \frac{4}{3} + \frac{1}{2}= \frac{8}{6} + \frac{3}{6} = \frac{11}{6}. \]

Step 6: Conclusion

The value of the double integral, which represents the area of the region of integration, is \( \frac{11}{6} \).

Key Concepts

Region of Integration

When evaluating a double integral, it's important to understand the 'region of integration'. This is the area over which we are calculating the integral.
The description of this region is given by the limits of integration for the variables involved.
In our exercise, we integrate over

• a variable region defined by the limit for \( x \) dependent on the value of \( y \).
• Specifically, \( x \) ranges between \( y \) and \( y^2 \) for each \( y \) in the interval from 1 to 2.

This means our region lies between the line \( x = y \) and the curve \( x = y^2 \).
A helpful way to visualize this is by sketching these boundaries on a coordinate plane.
The line \( x = y \) is straight, while \( x = y^2 \) forms a parabola that opens to the right.
The section we're interested in is therefore bounded by these curves, running from \( y = 1 \) to \( y = 2 \).
By shading this area on a graph, we can clearly see the region over which our double integral is calculated.

Evaluation of Integrals

In evaluating double integrals, we deal with multiple layers of integration.
First, we solve the innermost integral, treating the outer variable as a constant.
In our example, the inner integral is

• \( \int_{y}^{y^2} 1 \, dx \), where the integrand is simply 1.

Solving this integral with respect to \( x \), we find:

• \([x]_{y}^{y^2} = y^2 - y \).

This result is a function of \( y \) because we integrated with respect to \( x \).
Next, we insert this result into the outer integral,

• \( \int_{1}^{2} (y^2 - y) \, dy \),

and solve with respect to \( y \):

• \( \frac{y^3}{3} - \frac{y^2}{2} \big|_{1}^{2} \).

Evaluating this expression from \( y = 1 \) to \( y = 2 \) gives us the final solution, simplifying to \( \frac{11}{6} \).
This evaluated value represents a cumulative "sum" over the entire region defined by our limits of integration.

Integral Calculus

Integral calculus is a central part of mathematics dealing with the concept of integration.
It involves finding accumulative quantities, like area under a curve or total volume.

• In essence, an integral uses small parts or slices to compute a value over a certain range.

There are different types of integrals, and our focus here is on double integrals.
Double integrals extend the idea of an integral to functions of two variables, commonly \( x \) and \( y \).
They calculate the volume under a surface in a 3D space and are written as \( \int \int f(x, y) \, dx \, dy \).
This process requires integrating twice: once for each variable.

• Our example, \( \int_{1}^{2} \int_{y}^{y^2} d x d y \), keeps the integrand the same for simplicity, which simplifies the computations.

Understanding how integrals work allows us to solve real-world problems wherever quantities are accumulated over an area or volume.
Through this exercise, integral calculus helps us understand spatial shapes, giving a basis for mathematical modeling in physics, engineering, and other sciences.
Grasping these integrals is essential for a deeper comprehension of how changes occur across different dimensions.