Question

Use double integrals to find the indicated volumes. Under the surface \(z=1 /(y+2),\) and over the area bounded by \(y=x\) and \(y^{2}+x=2\).

Short answer

The volume is obtained by evaluating the integral \(\int_{0}^{1} \ln \left( \frac{\sqrt{2-x}+2}{x+2} \right) dx\).

Step-by-step solution

Understand the Region of Integration

First, identify the region in the xy-plane over which we need to integrate. The curves given are: \(y = x\) and \(y^2 + x = 2\). Find their points of intersection by setting \(y = x\) in the equation \(y^2 + x = 2\).

Find the Intersection Points

Solve \(x^2 + x = 2\). The solutions are \(x^2 + x - 2 = 0\). Factoring gives \((x-1)(x+2) = 0\), so \(x = 1, -2\). Since \(y = x\), the intersection points are \((1,1)\) and \((-2, -2)\). However, \(y = x\) and \(y^2 + x = 2\) do not intersect at (-2,-2) in the relevant domain, thus the correct region is bounded between (0, 0) and (1, 1).

Set Up the Double Integral

The integral changes of \(y = x\) to \(y = \sqrt{2-x}\). Setup the double integral with the bounds as follows: \(\int_{0}^{1} \int_{x}^{\sqrt{2-x}} \frac{1}{y+2} \, dy \, dx\). The integral bounds for \(y\) go from \(y = x\) to \(y = \sqrt{2-x}\), and for \(x\) is from 0 to 1.

Integrate with Respect to y

Evaluate the inner integral \(\int_{x}^{\sqrt{2-x}} \frac{1}{y+2} \, dy\). Use the substitution \(u = y + 2\) which gives \(du = dy\), and the bounds change from \(x+2\) to \(\sqrt{2-x}+2\). Therefore, the integral becomes \(\int_{x+2}^{\sqrt{2-x}+2} \frac{1}{u} \, du\), which simplifies to \(\ln|u|\).

Evaluate the Inner Integral

Compute \(\left[ \ln |u| \right]_{x+2}^{\sqrt{2-x}+2}\). Substitute back the limits to obtain: \(\ln (\sqrt{2-x} + 2) - \ln (x+2)\). Simplify the logarithm difference to get: \(\ln \left( \frac{\sqrt{2-x}+2}{x+2} \right)\).

Integrate with Respect to x

Set up the outer integral: \(\int_{0}^{1} \ln \left( \frac{\sqrt{2-x}+2}{x+2} \right) \, dx\). This integral is solved through numerical integration or a suitable substitution technique.

Compute the Final Value

Evaluate the definite integral using integration methods or computational tools. The exact value might not be easily expressible in a closed form, but numerical approximation methods can be utilized to find the volume under the surface.

Key concepts

region of integration

To tackle double integrals, first, understand the **region of integration**. This is the part of the xy-plane where we will evaluate our integral. In our example, we're given two curves: the line defined by \(y = x\) and the curve \(y^2 + x = 2\). To identify the region between these two curves, we need to find their points of intersection. Substituting \(y = x\) into \(y^2 + x = 2\), we solve for \(x^2 + x - 2 = 0\). Factoring this, we get \((x-1)(x+2)=0\), leading to the solutions \(x=1\) and \(x=-2\). Checking the relevant domain, the interesting region is from (0, 0) to (1, 1) where these two curves intersect within our integration bounds.

bounds of integration

After understanding the region of integration, set the **bounds of integration**. These bounds will determine the limits within which we are evaluating the double integral. For our problem, the integral bounds for \(y\) span from the line \(y = x\) to the curve \(y = \sqrt{2-x}\). Consequently, the bounds for \(x\) go from 0 to 1. Therefore, the double integral can be expressed as: \[ \int_{0}^{1} \int_{x}^{\sqrt{2-x}} \frac{1}{y+2} \ dy \ dx \]. Properly identifying these bounds is crucial since they define the region over which we calculate the volume.

numerical integration

Sometimes integrals, especially double integrals, don't have simple analytic solutions. When this happens, we can use **numerical integration** to approximate the value. In our case, the outer integral \( \int_{0}^{1} \ln\left( \frac{\sqrt{2-x}+2}{x+2} \right) \ dx \) may be difficult to solve analytically. Numerical methods such as the Trapezoidal Rule, Simpson's Rule, or numerical algorithms in computational tools (like MATLAB or Python's SciPy library) can be used to approximate this value. These methods work by summing up small slices under the curve, providing an approximation of the area.

substitution technique

The **substitution technique** is another helpful method in solving integrals by making them simpler. Here we use the substitution \(u = y + 2\) which transforms our inner integral into a more manageable form. Substituting, we get \(du = dy\), and our bounds change accordingly to \(x + 2\) and \(\sqrt{2 - x} + 2\). The transformed integral then becomes \[ \int_{x+2}^{\sqrt{2-x}+2} \frac{1}{u} \ du \]. This simplifies to \(\left[ \ln |u| \right]_{x+2}^{\sqrt{2-x}+2}\). Applying these techniques can help break down complex integrals into easier parts.

volume under surface

The final goal of using double integrals in this context is to find the **volume under the surface** defined by the function between the given regions and boundaries. In our context, the volume under the surface \(z = \frac{1}{y+2}\) above the region bounded by \(y = x\) and \(y^2 + x = 2\) is calculated using double integrals. Setting up correctly and solving the integral: \[ \int_{0}^{1} \int_{x}^{\sqrt{2-x}} \frac{1}{y+2} \ dy \ dx \] enables us to find this volume. Sometimes the final step might require numerical methods to approximate, but it serves as a definite means to find the volume under complex surfaces.