Question

Der Bereich \(B\) aus dem 1. Quadranten \((x \geq 0, y \geq 0)\) des \(\mathbb{R}^{2}\) wird durch die Kurven $$ y=3-x^{2}, \quad y=x^{2}+1 \quad \text { und } \quad x=0 $$ begrenzt. Berechnen Sie das Integral $$ \int_{B} \frac{x}{\sqrt{y}} d F $$

Short answer

Calculate the two integrals using substitution and combine them to find the total integral over the region.

Step-by-step solution

Understand the Problem

The region \( B \) is defined in the first quadrant and is bounded by the curves \( y = 3 - x^2 \), \( y = x^2 + 1 \), and the line \( x = 0 \). We need to find the integral \( \int_{B} \frac{x}{\sqrt{y}} dF \), where \( dF \) represents the differential area element in this region.

Identify the Intersection Points

Find where the curves \( y = 3 - x^2 \) and \( y = x^2 + 1 \) intersect. Set them equal: \\[ 3 - x^2 = x^2 + 1. \] \Solving for \( x \), we get \\[ 2 = 2x^2 \quad \Rightarrow \quad x^2 = 1 \quad \Rightarrow \quad x = 1 \text{ (since } x \geq 0). \] \The intersection point is \( (1, 2) \).

Describe the Integration Region

The region \( B \) is bounded vertically from \( x = 0 \) to \( x = 1 \). For each fixed \( x \) in this range, \( y \) varies from the lower curve \( y = x^2 + 1 \) to the upper curve \( y = 3 - x^2 \).

Set Up the Double Integral

Set up the double integral to integrate with respect to \( y \) first and then with respect to \( x \): \\[ \int_{0}^{1} \int_{x^2 + 1}^{3 - x^2} \frac{x}{\sqrt{y}} \, dy \, dx. \]

Integrate with Respect to \( y \)

Compute the inner integral: \\[ \int_{x^2 + 1}^{3 - x^2} \frac{x}{\sqrt{y}} \, dy. \] \To solve this, let \( u = \sqrt{y} \Rightarrow du = \frac{1}{2\sqrt{y}} \, dy \Rightarrow dy = 2u \, du \). Then, the integral becomes: \\[ 2x \int_{\sqrt{x^2 + 1}}^{\sqrt{3 - x^2}} du = 2x[u]_{\sqrt{x^2+1}}^{\sqrt{3-x^2}} = 2x(\sqrt{3-x^2} - \sqrt{x^2+1}). \]

Integrate with Respect to \( x \)

Integrate the resulting expression with respect to \( x \): \\[ \int_{0}^{1} 2x(\sqrt{3-x^2} - \sqrt{x^2+1}) \, dx. \] \This can be solved by splitting it into two integrals: \\[ 2\int_{0}^{1} x \sqrt{3-x^2} \, dx - 2\int_{0}^{1} x \sqrt{x^2+1} \, dx. \] \These integrals can be solved using substitution: for the first, let \( u = 3-x^2 \), and for the second, let \( v = x^2+1 \).

Solve and Evaluate Each Integral

Using the substitutions, calculate each integral independently. For example, for the first integral, the substitution \( u = 3-x^2 \) yields \\[ dx = -\frac{du}{2x} \] \which simplifies the manipulation. Solve both integrals to get specific numerical results for each to complete the integration.

Calculate Final Result

Combine the results from Step 7 to find the value of the original integral over the region \( B \). Compute the sum of the results of the solved integrals from Step 7.

Key concepts

Integration Techniques

Understanding the various integration techniques is crucial for solving double integrals, especially over a specified region. In this exercise, a double integral is used to evaluate \[ \int_{B} \frac{x}{\sqrt{y}} dF \] over a specific region \( B \). Integration is carried out over two variables: first with respect to \( y \), then \( x \). Different techniques include:

• Separation of Integrals: The problem can often be reduced to simpler single-variable integrals. This makes calculations more manageable and is demonstrated when solving the given integral in steps.


• Substitution Method: Substitution is employed when direct integration is challenging, such as in the inner integral using substitutions like \( u = \sqrt{y} \) to simplify the expression.


• Splitting and Summing Integrals: Breaking down the integral into manageable pieces by incorporating different limits of integration or by recognizing composite regions.

Mastering these techniques allows students to evaluate complex integrals, even when the functions involved aren’t straightforward.

Region of Integration

Defining the region of integration is a foundational step in solving integrals over specified areas. In this exercise, region \( B \) is delineated by curves and lines in the first quadrant of the plane, namely:

• The curve \( y = 3 - x^2 \)


• The curve \( y = x^2 + 1 \)


• The line \( x = 0 \)

These provide the boundaries within which integration occurs.
For effective integration, it's essential to understand the limits for each variable in \( B \):

• Horizontal limits (for \( x \)): Extend from \( x = 0 \) to the intersection point \( x = 1 \).


• Vertical limits (for \( y \)): For each fixed \( x \) in this horizontal range, \( y \) moves between \( x^2 + 1 \) and \( 3 - x^2 \).

Visualizing or sketching these boundaries helps in setting up the integration limits for the calculation of area or volume.

Intersection Points

Finding intersection points is essential to accurately define the region of integration. In our problem, curves intersect at points which help bound the region of \( B \) more precisely. To find these points, you solve the equations where curves meet. Here's how:

• The curves are \( y = 3 - x^2 \) and \( y = x^2 + 1 \). Setting them equal gives us \( 3 - x^2 = x^2 + 1 \).


• Rearranging terms yields \( 2 = 2x^2 \) and solving that gives \( x^2 = 1 \), so \( x = 1 \) due to \( x \geq 0 \).

Substituting \( x = 1 \) back into either equation for \( y \) provides the point \( (1, 2) \).
Identifying these intersection points allows us to establish where the curves intercept, aiding us in setting precise limits for the double integral's evaluation.

Substitution Method

At times, the integrand's complexity necessitates the substitution method to make integration feasible. This approach simplifies integration by changing variables to reduce complexity. In this step-by-step solution, substitution is employed while integrating with respect to \( y \):

• Given the integral \( \int \frac{x}{\sqrt{y}} \, dy \), the substitution approach calls for setting \( u = \sqrt{y} \). This transforms \( y \) into \( u^2 \) and \( dy = 2u \, du \).


• The integral simplifies to \( 2x \int du = 2x[u] \), making it simpler to evaluate over the adjusted limits.

Additionally, substitution in the final steps can further break down otherwise intricate expressions.
This method's main advantage lies in transforming difficult integrals into simpler, more standard forms, making them solvable through basic calculus rules.