Problem 23 - Calculus Early Transcendentals

Question

Sketch each region (if a figure is not given) and find its area by integrating with respect to \(y\) The region bounded by \(y=\sqrt{\frac{x}{2}+1}, y=\sqrt{1-x}\), and \(y=0\)

Step-by-Step Solution

Answer

Answer: \(\frac{2}{3\sqrt{3}}\)

Step 1: Sketch the given region

First, we will sketch the region bounded by the three curves \(y=\sqrt{\frac{x}{2}+1}\), \(y=\sqrt{1-x}\), and \(y=0\). To do this, divide this step into the following sub-steps: 1. Sketch the graph of \(y=\sqrt{\frac{x}{2}+1}\), which is a square root function shifted horizontally and vertically. 2. Sketch the graph of \(y=\sqrt{1-x}\), which is another square root function, reflecting the graph horizontally. 3. Sketch the graph of \(y=0\), which is a horizontal line along the x-axis. Observe that the bounded region is a closed area between these curves where all three functions are non-negative.

Step 2: Find the intersection points between the curves

To find the intersection points between the curves, we must solve the following equations for \(x\) and \(y\). 1. Intersection between \(y=\sqrt{\frac{x}{2}+1}\) and \(y=\sqrt{1-x}\): \[\sqrt{\frac{x}{2}+1} = \sqrt{1-x}\] Square both sides to get rid of the square roots: \[\frac{x}{2}+1=1-x\] Solve for \(x\): \[x=2-2x\] \[3x=2\] \[x=\frac{2}{3}\] Plug this value of \(x\) back into either equation to find the corresponding \(y\) value: \[y=\sqrt{1-\frac{2}{3}}\] \[y=\frac{1}{\sqrt{3}}\] So, the intersection point is \(\left(\frac{2}{3},\frac{1}{\sqrt{3}}\right)\). 2. Intersection between \(y=\sqrt{\frac{x}{2}+1}\) and \(y=0\): \[\sqrt{\frac{x}{2}+1}=0\] Squaring both sides: \[\frac{x}{2}+1=0\] Solve for \(x\): \[x=-2\] Plug this value of \(x\) back into the equation to find the corresponding \(y\) value: \[y=0\] So, the intersection point is \((-2,0)\). 3. Intersection between \(y=\sqrt{1-x}\) and \(y=0\): \[\sqrt{1-x}=0\] Squaring sides: \[1-x=0\] Solve for \(x\): \[x=1\] Plug this value of \(x\) back into the equation to find the corresponding \(y\) value: \[y=0\] So, the intersection point is \((1,0)\).

Step 3: Integrate with respect to \(y\)

Now that we have the points of intersection, we can integrate the curves with respect to \(y\). We'll integrate the difference between the two curve equations which involves solving for \(x\) in terms of \(y\): 1. Solve \(y=\sqrt{\frac{x}{2}+1}\) for \(x\): \[x=2(y^2-1)\] 2. Solve \(y=\sqrt{1-x}\) for \(x\): \[x=1-y^2\] Now, set up the integration to find the area of the region: Area = \(\int_{y=0}^{y=\frac{1}{\sqrt{3}}} [(2(y^2-1))-(1-y^2)]dy\) Simplify the integrand: Area = \(\int_{y=0}^{y=\frac{1}{\sqrt{3}}} (3y^2-1)dy\) Find the antiderivative: Area = \(\left[ y^3 - y \right]_{y=0}^{y=\frac{1}{\sqrt{3}}}\) Evaluate the definite integral: Area = \(\left[\left(\frac{1}{\sqrt{3}}\right)^3 - \left(\frac{1}{\sqrt{3}}\right)\right] - [0^3-0]\) Area = \({\frac{1}{3\sqrt{3}}-\frac{1}{\sqrt{3}}}\) Area = \({\frac{-2}{3\sqrt{3}}}\) The area of the region bounded by the curves is \({\frac{-2}{3\sqrt{3}}}\). However, since the area cannot be negative, we take the absolute value to get: Area = \({\frac{2}{3\sqrt{3}}}\)

Key Concepts

Definite Integral

When we speak of finding the area between curves, we rely on a powerful mathematical tool known as the definite integral. The definite integral, notated as \[\int_{a}^{b} f(x)dx\], represents the accumulation of quantities, which can be interpreted as the area under the curve of a function from one point on the x-axis to another.

In the context of area calculations, we're often dealing with the definite integral of the vertical 'slices' of space between two curves, from a starting point 'a' to an ending point 'b'. The value of this integral tells us the 'net' area - considering the portions above the x-axis as positive and those below as negative. Therefore, if our definite integral yields a negative number, we take its absolute value to get the actual area, since area cannot be negative.

Integration with Respect to y

Traditionally, definite integrals are calculated with respect to the variable x. However, when the functions are better described as functions of y (for example, when they are more vertical or involve a square root of x), it can be beneficial to integrate with respect to y. This approach flips the perspective, considering horizontal 'slices' instead of vertical ones, and integrating along the y-axis.

The process involves solving the given curve equations for x as a function of y, setting up the integral with y-limits from the lower bound to the upper bound of the region, and integrating the difference of x-values corresponding to these bounds. This is precisely what was done in our example, where the area between the curves was found using this method.

Intersection of Curves

Finding the intersection of curves is an essential step in determining the area between them because intersections represent the limits of integration. The intersection points specify where the curves meet, which in essence frames the region whose area we’re interested in. We often find these points by setting the equations of the curves equal to each other and solving for the variables involved.

In our example, the intersections were found by equating the two given square root functions and solving for x and y, which gave us the bounds needed for the integration. These intersection points help delineate the region of interest, ensuring that the area calculation is accurate for the specified section of the graph.