Problem 47 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Use a double integral to find the area of \(R\).\(R\) is the region bounded by \(y=\frac{1}{2} x^{2}\) and \(y=2 x\).

Step-by-Step Solution

Answer

The area of region \( R \) is \( \frac{16}{3} \).

Step 1 - Determine the Points of Intersection

To find the bounds of integration, calculate where the curves intersect by setting the equations equal to each other: \(\frac{1}{2} x^2 = 2x\). Solve for \(x\) to find the points of intersection.

Step 2 - Solve the Equation

Solve the equation from Step 1: \(\frac{1}{2} x^2 = 2x\) Rearrange it to: \(\frac{1}{2} x^2 - 2x = 0\) Factor out \(x\): \(x(\frac{1}{2} x - 2) = 0\) So, \(x = 0\) or \(x = 4\). Thus, the points of intersection are \((0, 0)\) and \((4, 8)\).

Step 3 - Set Up the Double Integral

Since the region \( R \) is bounded between the curves \( y = \frac{1}{2} x^2 \) and \( y = 2x \), set up the double integral for the area: \[A = \int_{0}^{4} \int_{\frac{1}{2} x^2}^{2x} dy \, dx\]

Step 4 - Evaluate the Inner Integral

Evaluate the inner integral with respect to \(y\): \[\int_{\frac{1}{2} x^2}^{2x} dy = [y]_{\frac{1}{2} x^2}^{2x} = 2x - \frac{1}{2} x^2\]

Step 5 - Evaluate the Outer Integral

Now, evaluate the outer integral: \[\int_{0}^{4} (2x - \frac{1}{2} x^2) dx\] Split it into two integrals: \[\int_{0}^{4} 2x \, dx - \int_{0}^{4} \frac{1}{2} x^2 \, dx\] These integrals are straightforward to solve.

Step 6 - Solve the Outer Integrals

Solve the integrals from Step 5: \[\int_{0}^{4} 2x \, dx = [x^2]_{0}^{4} = 16 \]\[\int_{0}^{4} \frac{1}{2} x^2 \, dx = [\frac{1}{6} x^3]_{0}^{4} = \frac{32}{3}\]

Step 7 - Find the Area

Subtract the results from Step 6 to find the area: \[16 - \frac{32}{3} = \frac{48}{3} - \frac{32}{3} = \frac{16}{3}\] Therefore, the area of region \( R \) is \( \frac{16}{3} \).

Key Concepts

area between curves

To find the area between two curves using double integrals, we first need to understand how these curves form a region. For our exercise, we have two curves: \(y = \frac{1}{2} x^2\) and \(y = 2x\). The area between these curves is the region enclosed by these equations.

• Imagine drawing both curves on a coordinate plane.
• The area we are interested in is the space where one curve is above the other, within specific bounds.

This area can be found by calculating a double integral. Integrals let us sum up infinitesimally small vertical slices of the area between these curves, from the leftmost point to the rightmost point where they intersect.

points of intersection

To set up our integral correctly, we first need to identify the points where the two curves intersect. This gives us the limits of integration.

• Set the equations equal to each other: \( \frac{1}{2} x^2 = 2x \).
• Solve for \(x\).
• Rearrange to \( \frac{1}{2} x^2 - 2x = 0 \).
• Factor out \(x\) to get \ x(\frac{1}{2} x - 2) = 0 \.

This gives \( x = 0 \) and \( x = 4 \) as the points of intersection. The corresponding \(y\)-values are obtained by substituting back into either equation: \( (0, 0) \) and \( (4, 8) \). These points are the boundaries of our region.

calculus integrals

With our points of intersection known, we can now set up and solve our integrals.

• First, set up the double integral: \ A = \int_{0}^{4} \int_{\frac{1}{2} x^2}^{2x} dy \, dx \.
• We evaluate the inner integral with respect to \(y\), \ \int_{\frac{1}{2} x^2}^{2x} dy = [y]_{\frac{1}{2} x^2}^{2x} = 2x - \frac{1}{2} x^2 \.

The next step is to evaluate the remaining outer integral with respect to \(x\): \ \int_{0}^{4} (2x - \frac{1}{2} x^2) dx \.
We can split this integral into two parts: \ \int_{0}^{4} 2x \ dx - \int_{0}^{4} \frac{1}{2} x^2 \ dx \.

bounding equations

Understanding the bounding equations is crucial. They define the limits within which we integrate.

• We are bounded by \(y = \frac{1}{2} x^2 \) from below and \( y = 2x \) from above.
• These boundaries determine the region, along with the vertical spans from \( x = 0 \) to \( x = 4 \).

The final step is to solve these outer integrals:

• Solve \int_{0}^{4} 2x \, dx = [x^2]_{0}^{4} = 16 \.
• Solve \ \int_{0}^{4} \frac{1}{2} x^2 \, dx = [\frac{1}{6} x^3]_{0}^{4} = \frac{32}{3} \.

Subtract the two results to find the area: \ 16 - \frac{32}{3} = \frac{48}{3} - \frac{32}{3} = \frac{16}{3}.
Therefore, the area of region \ R \ is \ \frac{16}{3} \.