Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the indicated double integral over \(R\). \(\iint_{R}\left(x^{2}+y^{2}\right) d A ; R=\\{(x, y):-1 \leq x \leq 1,0 \leq y \leq 2\\}\)

Step-by-Step Solution

Answer

The value of the double integral is \( \frac{20}{3} \).

Step 1: Set Up the Double Integral

We are given the region \( R = \{ (x, y) : -1 \leq x \leq 1, 0 \leq y \leq 2 \} \). The integrand is \( x^2 + y^2 \). Therefore, the double integral is set up as follows:\[\int_{0}^{2} \int_{-1}^{1} (x^2 + y^2) \, dx \, dy.\]

Step 2: Integrate with Respect to x

Perform the integration with respect to \( x \). Treat \( y^2 \) as a constant in this step:\[\int_{-1}^{1} (x^2 + y^2) \, dx = \int_{-1}^{1} x^2 \, dx + \int_{-1}^{1} y^2 \, dx.\]Calculate each of these integrals separately:\( \int_{-1}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-1}^{1} = \frac{1}{3} - (-\frac{1}{3}) = \frac{2}{3} \).For \( \int_{-1}^{1} y^2 \, dx = y^2 \cdot (x) \big|_{-1}^{1} = y^2 \cdot (1 - (-1)) = 2y^2 \).Thus, the integrated result is \( \frac{2}{3} + 2y^2 \).

Step 3: Integrate with Respect to y

Now we integrate the result from Step 2 with respect to \( y \):\[\int_{0}^{2} \left( \frac{2}{3} + 2y^2 \right) \, dy.\]Calculate each part separately:\( \int_{0}^{2} \frac{2}{3} \, dy = \frac{2}{3} \cdot y \big|_{0}^{2} = \frac{4}{3} \).\( \int_{0}^{2} 2y^2 \, dy = 2 \cdot \left[ \frac{y^3}{3} \right]_{0}^{2} = \frac{2(8)}{3} - 0 = \frac{16}{3} \).Add them together: \( \frac{4}{3} + \frac{16}{3} = \frac{20}{3}. \)

Step 4: Final Answer

The result of the double integral over the region \( R \) is \( \frac{20}{3} \). This is the total value of the integral \( \iint_{R} (x^2 + y^2) \, dA \) when integrated over the specified boundaries.

Key Concepts

Understanding the Integrand

In a double integral, the integrand is the function we are integrating over a given region. In this particular exercise, the integrand is given by the expression \(x^2 + y^2\). This expression contributes to the value of the overall integral as it accounts for different values of \(x\) and \(y\) inside the region of integration.
- **\(x^2\):** Represents the squared value of the \(x\)-coordinate, influencing the integral as \(x\) varies within its limits.- **\(y^2\):** Represents the squared value of the \(y\)-coordinate, treated as a constant during the integration with respect to \(x\), but varying when integrating with respect to \(y\).
The purpose of evaluating \(x^2 + y^2\) across the specified region is to calculate how these values accumulate, which in some contexts could relate to things like mass, area, or other physical properties.

Integration with Respect to x

When we perform integration with respect to \(x\), we treat \(y\) (and thus \(y^2\)) as a constant. This simplifies the integrand, allowing us to focus solely on the effects of \(x\).
- The integral becomes \(\int_{-1}^{1} (x^2 + y^2) \, dx\), breaking down into \(\int_{-1}^{1} x^2 \, dx\) and \(\int_{-1}^{1} y^2 \, dx\).- Integrating \(x^2\) results in \([\frac{x^3}{3}]_{-1}^1 = \frac{2}{3}\), capturing the change in height component due to the \(x\) variation.
- Integrating \(y^2\) as a constant factor over \(x\)'s range gives \(2y^2\), representing a simple scaling of \(y^2\) by the width of the interval \/([\-1, 1]\/).

Combining these results, we get \(\frac{2}{3} + 2y^2\), which we then use in the next integration step.

Integration with Respect to y

After integrating with respect to \(x\), we move on to integrating with respect to \(y\). This step consolidates the earlier result over the whole \(y\)-range
- The integral expression is \(\int_{0}^{2} (\frac{2}{3} + 2y^2) \, dy\), which we split into two parts: \(\int_{0}^{2} \frac{2}{3} \, dy\) and \(\int_{0}^{2} 2y^2 \, dy\).- Integrating \(\frac{2}{3}\) gives \(\frac{4}{3}\), derived from multiplying the constant by the length of the interval \([0, 2]\).
- For \(2y^2\), the integration \([\frac{y^3}{3}]_{0}^{2} = \frac{16}{3}\) comes from calculating the change across the \(y\)-values and scaling by 2.
By adding these contributions together, \(\frac{4}{3} + \frac{16}{3} = \frac{20}{3}\), we obtain the total value of the original double integral over the specified region.

Defining the Region of Integration

The region of integration is essential in setting the boundaries for our double integral. It specifies where we evaluate the integrand. In this exercise, the region \(R\) is defined as \((x, y) : -1 \leq x \leq 1, 0 \leq y \leq 2\).
- **\(-1 \leq x \leq 1\):** Indicates that we are only considering \(x\)-values between -1 and 1.
- **\(0 \leq y \leq 2\):** Restricts \(y\)-values to the range from 0 to 2.

This rectangular domain allows us to systematically integrate first with respect to \(x\) over this interval and then with respect to \(y\). Understanding and visualizing this region helps interpret the integral as essentially accumulating contributions from every point within this defined space, aligning perfectly with the integrand terms. Remember, the choice of limits affects our integral outcome, so these boundaries are crucial.