Problem 1 - Thomas Calculus

Question

In Exercises \(1-10,\) sketch the region of integration and evaluate the integral. $$ \int_{0}^{3} \int_{0}^{2}\left(4-y^{2}\right) d y d x $$

Step-by-Step Solution

Answer

The value of the integral is 16.

Step 1: Understand the Limits of Integration

The given double integral is \(\int_{0}^{3} \int_{0}^{2}(4-y^2) \, dy \, dx\). The inner integral (with respect to \(y\)) is over [0, 2], and the outer integral (with respect to \(x\)) is over [0, 3]. This describes a rectangular region in the xy-plane, where \(0 \leq x \leq 3\) and \(0 \leq y \leq 2\).

Step 2: Sketch the Region of Integration

Draw the xy-plane and mark the limits of integration for \(x\) from 0 to 3 and for \(y\) from 0 to 2. The region of integration is a rectangle with vertices at points (0,0), (3,0), (3,2), and (0,2).

Step 3: Integrate with Respect to y

Start by integrating the function \(4-y^2\) with respect to \(y\) over [0, 2].\[\int_{0}^{2} (4-y^2) \, dy = \left[ 4y - \frac{y^3}{3} \right]_{0}^{2}\]Substitute the limits:\[= \left(4\cdot 2 - \frac{2^3}{3}\right) - \left(4\cdot 0 - \frac{0^3}{3}\right)\]\[= 8 - \frac{8}{3} = \frac{24}{3} - \frac{8}{3} = \frac{16}{3}\]

Step 4: Integrate with Respect to x

Now, integrate the result of the first integral, \(\frac{16}{3}\), with respect to \(x\) over [0, 3]:\[\int_{0}^{3} \frac{16}{3} \, dx = \frac{16}{3} \times \left[ x \right]_{0}^{3}\]Substitute the limits:\[= \frac{16}{3} \times (3 - 0) = \frac{16}{3} \times 3 = 16\]

Step 5: Conclude the Evaluation

The value of the original double integral is 16, which represents the volume under the surface \(z = 4 - y^2\) over the rectangular region bounded by 0 \(\le x \le 3\) and 0 \(\le y \le 2\).

Key Concepts

Region of Integration

In integral calculus, the region of integration refers to the area over which we perform the integration. For double integrals, this is a two-dimensional space typically described in the xy-plane. In this particular exercise, the region is presented via the limits of integration.
A key aspect to understand is that these limits define a rectangular region.

• The inner integral specifies the limits for the variable \( y \), which in this case is from 0 to 2.
• The outer integral outlines the limits for \( x \), ranging from 0 to 3.

By sketching this region in the xy-plane, you'll notice it forms a rectangle with vertices at (0,0), (3,0), (3,2), and (0,2). It's crucial to visualize this to understand the boundaries of your integration.
Being adept at identifying and sketching these regions helps in setting up integrals correctly, paving the way for accurate calculations.

Integral Calculus

Integral calculus is a fundamental area of mathematics used to calculate quantities like areas, volumes, and sums. In the context of this problem, we employ a double integral, a tool that extends the concept of integration to two dimensions. This enables us to find volumes under surfaces over specific regions.The double integral \( \int_{0}^{3} \int_{0}^{2} (4 - y^2) \, dy \, dx \) is evaluated by performing two integrations: the inner with respect to \( y \), and the outer with respect to \( x \).

• First, we integrate the function \( 4 - y^2 \) with respect to \( y \), which simplifies the expression to evaluate over \( x \).
• Then, we perform the integration with respect to \( x \) using the resulting value to find the entire volume under the surface.

Integral calculus, especially through double integrals, provides the machinery to explore two-dimensional spaces and compute various physical and mathematical properties.

Volume Under a Surface

Finding the volume under a surface using double integration is an insightful application of integral calculus. In this problem, the surface is defined by the function \( z = 4 - y^2 \). This surface spans the rectangular region outlined by the limits of integration.

• The integration process involves "slicing" the volume into thin layers parallel to the xy-plane and summing them up.
• Each of these "slices" has a thickness, and integrating across them computes the overall volume.

After performing both integrations, the resultant value, in this case, is 16. This represents the total volume beneath the surface within the boundaries \( 0 \le x \le 3 \) and \( 0 \le y \le 2 \). Understanding how to compute such volumes is essential in physics and engineering to solve real-world problems involving 3D shapes and regions.