Problem 56 - Calculus Early Transcendentals

Question

Use double integrals to calculate the volume of the following regions. The solid beneath the cylinder \(z=y^{2}\) and above the region \(R=\\{(x, y): 0 \leq y \leq 1, y \leq x \leq 1\\}\)

Step-by-Step Solution

Answer

Answer: The volume of the solid is \(\frac{1}{36}\).

Step 1: Express the double integral

First, we need to express the double integral for the volume. The volume of the solid will be given by the integral of the function z = y^2 over the region R. We can write this integral as: $$ V = \iint_R y^2 dA $$

Step 2: Determine the limits of integration

Next, we need to determine the limits of integration for x and y. Given the region R: $$ 0 \leq y \leq 1 \\ y \leq x \leq 1 $$ We can see that y varies from 0 to 1 and x varies from y to 1. Now we can write the integral for the volume V as: $$ V = \int_{0}^{1} \int_{y}^{1} y^2 dxdy $$

Step 3: Evaluate the integral with respect to x

Now, we will first integrate the function with respect to x: $$ V = \int_{0}^{1} \left[ \frac{1}{3}y^2x^3 \right]_{y}^{1} dy \\ V = \int_{0}^{1} \frac{1}{3}y^2 (1^3 - y^3) dy $$

Step 4: Evaluate the integral with respect to y

Lastly, we will integrate the function with respect to y: $$ V = \left[ \frac{1}{12}y^5 - \frac{1}{18}y^8 \right]_{0}^{1} \\ V = \frac{1}{12}(1)^5 - \frac{1}{18}(1)^8 - \left(\frac{1}{12}(0)^5 - \frac{1}{18}(0)^8\right) \\ V = \frac{1}{12} - \frac{1}{18} $$

Step 5: Compute the volume

Finally, we will subtract the value we found in step 4 to get the volume: $$ V = \frac{1}{12} - \frac{1}{18} \\ V = \frac{3 - 2}{36} \\ V = \frac{1}{36} $$ So the volume of the solid beneath the cylinder z = y^2 and above the region R is \(\frac{1}{36}\).

Key Concepts

Volume Calculation

Calculating volume using double integrals is a powerful technique that extends the single-variable integral method into higher dimensions. In this case, we are working with a three-dimensional region beneath a surface and above a specified domain in the xy-plane.
To find the volume of the solid, we integrate the height function over the given region. The height function here is provided by the equation of the cylinder, which in this example is expressed as \( z = y^2 \). Consequently, the integral expression represents the sum of all infinitesimal prisms with base \( dA \) in the region \( R \) and height equal to \( y^2 \).
The double integral is set up to calculate the volume:\[ V = \iint_R y^2 \, dA \] Breaking it down, this involves two levels of integration: one for each dimension of the region \( R \). As a result, we evaluate this double integral in steps focusing on the iterated integral technique.

Limits of Integration

Determining the correct limits of integration is crucial for solving double integrals as they reflect the boundaries of the region of interest. In our example, we're given a specific region \( R \) in the xy-plane defined by the inequalities:

• \( 0 \leq y \leq 1 \)
• \( y \leq x \leq 1 \)

These inequalities define a triangular region within the first quadrant.
To establish the limits of integration for a double integral, we typically follow these steps:

• Identify the outer integral limits by addressing the total range of the first variable (here, \( y \) ranges from 0 to 1).
• Identify the inner integral limits by expressing the secondary variable (here \( x \)) based on the primary variable's current limits (from \( y \) to 1).

Bringing all these together, the integral can be expressed as:\[V = \int_{0}^{1} \int_{y}^{1} y^2 \, dx \, dy\]These limits ensure that the integral is evaluated over the entire triangular region \( R \).

Iterated Integrals

Iterated integrals allow us to evaluate double integrals by breaking the process down into sequential steps. This means we perform integration over one variable, followed by integration over a second variable. Each step refines the accumulated volume, adhering to the region's constraints.
For this exercise, first, we integrate with respect to \( x \), recognizing it as a placeholder while treating other variables as constants. Solving the integral:\[V = \int_{0}^{1} \left[ \frac{1}{3}y^2 x^3 \right]_{y}^{1} \, dy = \int_{0}^{1} \frac{1}{3} y^2 (1^3 - y^3) \, dy\]Next, the integral is completed by performing integration with respect to \( y \):\[V = \left[ \frac{1}{12}y^5 - \frac{1}{18}y^8 \right]_{0}^{1}\]Completing these integrations step-by-step consolidates the calculation of the full double integral. This methodical approach ultimately yields the final volume \( V = \frac{1}{36} \).
Beyond solving problems, iterated integrals are an invaluable tool for evaluating regions in various multidimensional spaces.