Problem 28 - Calculus an Applied Approach

Question

Sketch the region \(R\) whose area is given by the double integral. Then change the order of integration and show that both orders yield the same area. $$ \int_{0}^{4} \int_{0}^{\sqrt{x}} d y d x $$

Step-by-Step Solution

Answer

The area of region \(R\) obtained after evaluating the double integral and changing the order of integration for both cases is \(8\). So, both orders yield the same area for region \(R\).

Step 1: Sketch the region R

The region \(R\) is given by the limits of integration. In this case, \(0 \leq y \leq \sqrt{x}\), \(0 \leq x \leq 4\). If we sketch this on a graph, we'll see that the region \(R\) is bounded by the x-axis, the line \(y=\sqrt{x}\), and the vertical line \(x=4\).

Step 2: Evaluate the Double Integral

Next, we evaluate the given integral using the given boundaries \(\int_{0}^{4} \int_{0}^{\sqrt{x}} d y d x\). This gives \(1/2 x^{3/2 }\) between \(0\) and \(4\). It simplifies to \(8 \) units squared. This is the area of the region \(R\) obtained using the original order of integration.

Step 3: Change the Order of Integration

Now change the order of integration. We see that y ranges from 0 to 2 and x ranges from \(y^{2}\) to 4. So the new double integral becomes \(\int_{0}^{2}\int_{y^{2}}^{4} dxdy\).

Step 4: Prove the Areas are the Same

Evaluate the new double integral. This gives \(4y- y^{3}/3\) between \(0\) and \(2\) which also simplifies to \(8\). Both orders of integration yield the same area, proving that the statement is true.

Key Concepts

Order of Integration

When solving a double integral, the order of integration refers to the sequence in which you perform the integrations. In our original problem, the order is set as first integrating with respect to \(y\) and then \(x\), denoted by \(\int_{0}^{4} \int_{0}^{\sqrt{x}} dy \, dx\). Changing the order of integration often involves reconsidering the boundaries of integration to ensure they accurately describe the region. This change does not affect the outcome, but it may simplify the integration process or provide better insight into the problem.

• Original Order: \(dy \, dx\)
• New Order: \(dx \, dy\)

It is crucial to ensure that the region of integration remains the same when changing the order, as demonstrated by evaluating the integral under both orders to confirm the equal areas.

Area Calculation

The primary purpose of a double integral in this context is to calculate the area of the region \(R\). The problem involves finding the integral of 1 over the region, which effectively sums up all the infinitesimally small rectangles in the region. The calculation is straightforward once the limits of integration are set, as they direct you across the span of the desired region.Using the given boundaries, the area for our specific problem is:\[\int_{0}^{4} \int_{0}^{\sqrt{x}} dy \, dx \]Upon solving this integral, you should find the result is \(8\) square units, indicative of the complete area within the specified bounds. This single value represents the total area of the region defined by the limits and ensures confidence in the integration process's accuracy.

Integral Boundaries

Integral boundaries are the 'limits' or the endpoints for your integral. They help you define the region over which you are integrating. In the given double integral, the boundaries describe a specific part of the plane:- \(y\) ranges from \(0\) to \(\sqrt{x}\) - \(x\) ranges from \(0\) to \(4\)Changing the boundaries is crucial when altering the order of integration. After adjusting, the limits define where \(y\) can start and stop (0 to 2), and how \(x\) varies with respect to \(y\) (from \(y^2\) to 4). This adjustment ensures the same region is integrated upon when changing the order of integration.

Graphical Representation of Regions

Visualizing the region of integration is a practical step in understanding double integrals. Sketching the region, based on the given limits of integration, allows you to grasp the boundaries and the form of the region. For instance, the region here is defined by:

• The x-axis\( (y=0)\)
• The function \( y=\sqrt{x} \)
• The vertical line \( x=4 \)

By drawing these, you can see the shape is bounded below by the x-axis, above by the curve, and vertically at \(x=4\). When the order is changed to \(dx \, dy\), the region stays the same, though the interpretation of limits shifts. Visualization is key, as it verifies your limits and provides a clear geometric perspective on the problem.