Question

Sketch the regions of integration and evaluate the following integrals. \(\iint_{R} x^{2} y d A ; R\) is bounded by \(y=0, y=\sqrt{x},\) and \(y=x-2\)

Short answer

In summary, we have calculated the double integral of \(x^2y\) over the triangular region R formed by the curves \(y=0\), \(y=\sqrt{x}\), and \(y=x-2\). By sketching the region and setting up the double integral, we determined that: $$\iint_{R} x^{2} y dA = \frac{315}{4}$$

Step-by-step solution

Sketch the region and find its boundaries.

To identify the region R, sketch the three curves: 1. \(y=0\) is the x-axis. 2. \(y=\sqrt{x}\) is the curve that represents the boundary where the x and y values are the same. 3. \(y=x-2\) is a straight line that goes through points (2, 0) and (0, -2). To find the point of intersection between the curve \(y=\sqrt{x}\) and the line \(y=x-2\), set the two equations equal to each other: \(\sqrt{x} = x-2\) Square both sides: \(x = (x-2)^2\) Expand and solve the quadratic equation: \(x = x^2 - 4x + 4\) \(x^2 - 5x + 4 = 0\) Factor the equation: \((x-1)(x-4)=0\) So, x = 1 or x = 4. Now we have two points of intersection: (1, 1) and (4, 2). Since the region is bounded by the x-axis on the bottom, \(y=0\), the curve \(y=\sqrt{x}\) on left and the line \(y=x-2\) on right, the region of integration R is the triangular region formed by the points (1, 0), (1, 1) and (4, 2). Now that we've identified the region, it's time to set up the integral.

Set up the double integral.

It is more convenient to use the dy dx order. Find the limits of integration for y and x, taking into account the order of integration. Looking at the triangular region: - For a given x between 1 and 4, the y-value changes between the curve \(y=\sqrt{x}\) and the straight line \(y=x-2\). - The x-value goes from 1 to 4. So, we have: $$\iint_{R} x^{2} y dA = \int_1^4 \int_{\sqrt{x}}^{x-2} x^2 y dy dx$$ Now it's time to evaluate the integral.

Evaluate the inner integral.

We first evaluate the inner integral with respect to y: Inner integral: \(\int_{\sqrt{x}}^{x-2} x^2 y dy\) Integrate with respect to y: \(\frac{1}{2}x^2y^2 \Big|_{\sqrt{x}}^{x-2}\) Evaluate at the limits: \(\frac{1}{2}x^2 [(x-2)^2 - x]\) Now we will evaluate the outer integral.

Evaluate the outer integral.

Now, evaluate the outer integral with respect to x: Outer integral: \(\int_1^4 \frac{1}{2}x^2[(x-2)^2-x] dx\) Expand the equation and integrate term by term: \(\frac{1}{2}\int_1^4 (x^5-4x^4+8x^3-4x^2) dx\) Now integrate: \(\frac{1}{2}[\frac{1}{6}x^6 - \frac{4}{5}x^5 + 2x^4 - \frac{4}{3}x^3] \Big|_1^4\) Evaluate at the limits: \(\frac{1}{2}[(\frac{1}{6}(4^6) - \frac{4}{5}(4^5) + 2(4^4) - \frac{4}{3} (4^3)) - (\frac{1}{6}(1^6) - \frac{4}{5}(1^5) + 2(1^4) - \frac{4}{3}(1^3))]\) After calculating, we get: $$\iint_{R} x^{2} y dA = \frac{315}{4}$$ The value of the integral is \(\frac{315}{4}\).

Key concepts

Region of Integration

In this exercise, the region of integration is defined by three curves: \(y = 0\), \(y = \sqrt{x}\), and \(y = x - 2\). Understanding this region is crucial for solving the double integral.
The region forms a triangle with specific boundaries that limit the area we are integrating over. These curves outline the shape within which we calculate the integral.
- **\(y = 0\)** is simply the x-axis.- **\(y = \sqrt{x}\)** outlines a curve opening sideways.- **\(y = x - 2\)** is a straight line that defines another boundary.
The intersections of these lines help in precisely defining the region. The intersection points here are found at \((1, 1)\) and \((4, 2)\), forming a triangular integration area. By sketching these curves, you can visualize how they meet and enclose a specific region on the plane.

Boundaries in Integration

Determining the correct boundaries is a fundamental step in solving double integrals. These boundaries dictate the limits of your integrations both for \(x\) and \(y\).
To decide the boundaries:

• For \(y\), it varies between two curves: from \(\sqrt{x}\) to \(x - 2\).
• For \(x\), it ranges from 1 to 4, as determined by the intersection points.

By carefully choosing these limits, you ensure the double integral only considers the designated region. This setup means that we first integrate with respect to \(y\), followed by \(x\), capturing the correct part of the triangular area.

Calculus Integration Techniques

With the region and boundaries identified, we apply integration techniques to evaluate the integral. Double integration is performed in steps: first an inner integral and then the outer.
The inner integral is evaluated with respect to \(y\):

• This involves integrating \(x^2 y\) with the limits \(\sqrt{x}\) to \(x - 2\).

Next, the outer integral is done with respect to \(x\):

• Here, you evaluate the expression from the inner integral over \(x\) ranging from 1 to 4.

Each integration requires careful attention to detail, often involving substitution or breaking a complex polynomial into simpler parts to integrate term by term. These calculus techniques finally give us the integral's solution, allowing us to calculate the area under the curve precisely.