Question

Draw the solid region whose volume is given by the following double integrals. Then find the volume of the solid. $$\int_{0}^{1} \int_{-1}^{1}\left(4-x^{2}-y^{2}\right) d x d y$$

Short answer

To summarize: 1. We identified the limits of integration, which showed we are integrating over a rectangular region in the xy-plane: 0 ≤ y ≤ 1 and -1 ≤ x ≤ 1. 2. We recognized the function being integrated as a downward-opening paraboloid with a vertex at (0,0,4) and axis aligned with the z-axis. 3. We visualized the solid region as a portion of the paraboloid within the given rectangle in the xy-plane. 4. We calculated the volume of the solid region by evaluating the given double integral, resulting in a volume of \(\frac{16}{3}\) cubic units.

Step-by-step solution

Identify the limits of integration

The limits of integration are given in the double integral itself: $$\int_{0}^{1} \int_{-1}^{1}\left(4-x^{2}-y^{2}\right) d x d y$$ The outer integral limits are from 0 to 1, meaning \(0 \leq y \leq 1\). The inner integral limits are from -1 to 1, meaning \(-1 \leq x \leq 1\). These limits show that we are integrating the function over a rectangular region in the xy-plane, with 0 ≤ y ≤ 1 and -1 ≤ x ≤ 1.

Recognize the function being integrated

The function being integrated in this double integral is: $$4 - x^2 - y^2$$ This is a downward-opening paraboloid with the vertex at (0,0,4) and axis aligned with the z-axis.

Visualize the solid region

Using the limits of integration and the function from the previous steps, we can visualize the solid region as a portion of a downward-opening paraboloid with its vertex at (0,0,4) and axis aligned with the z-axis. The region is sliced along the xy-plane within the rectangle determined by the limits of integration for x and y. The solid region lies above the xy-plane (since the integrand is always positive within the given limits), within the given rectangle.

Calculate the volume of the solid region

To find the volume of the solid region, we'll evaluate the given double integral: $$ V = \int_{0}^{1} \int_{-1}^{1}\left(4-x^{2}-y^{2}\right) d x d y = \int_{0}^{1} \left(\int_{-1}^{1}\left(4-x^{2}-y^{2}\right) d x\right) d y$$ First, we integrate with respect to x: $$ \int_{-1}^{1} (4 - x^2 - y^2) dx = \left[4x - \frac{x^3}{3} - y^2x\right]_{-1}^1$$ Evaluate the antiderivative at the limits of integration: $$ \left[(4(1) - \frac{1^3}{3} - y^2(1))\right] - \left[(4(-1) - \frac{(-1)^3}{3} - y^2(-1))\right]$$ $$ = 4\left( \left(1 - y^2\right) + \left( 1-y^2 \right) \right)$$ Now, we have a single integral with respect to y: $$ V = \int_{0}^{1} 8(1 - y^2) dy$$ Integrate with respect to y: $$ V = \left[ 8y - \frac{8y^3}{3}\right]_0^1 $$ Evaluate the antiderivative at the limits of integration: $$ V = \left[ 8(1) - \frac{8(1)^3}{3}\right] - \left[ 8(0) - \frac{8(0)^3}{3}\right] $$ $$ V = 8 - \frac{8}{3} = \frac{16}{3} $$ The volume of the solid region is \(\frac{16}{3}\) cubic units.

Key concepts

Limits of Integration

In the realm of calculus, particularly when dealing with double integrals, it's crucial to grasp the concept of limits of integration. These limits define the region over which we integrate our function. For the integral provided, \[ \int_{0}^{1} \int_{-1}^{1} \left(4-x^{2}-y^{2}\right) \, dx \, dy\] the limits for \(x\) range from -1 to 1, and for \(y\) from 0 to 1.
These boundaries describe a rectangle in the \(xy\)-plane between \(-1 \leq x \leq 1\) and \(0 \leq y \leq 1\).
This rectangle serves as the base for the solid shape we are analyzing. Defining these limits accurately is the foundation of solving double integrals effectively, as it sets constraints on the region over which the function will be integrated.

Paraboloid

The function at the heart of our integral, \( 4-x^{2}-y^{2} \), resembles the equation of a paraboloid. This particular shape is a type of surface that curves downward from a single point, known as the vertex. In this equation, the paraboloid opens downwards with its vertex positioned at \((0,0,4)\) and the axis aligned along the \(z\)-axis.
Mathematically, a paraboloid can be expressed in the standard form \( z = a - x^2 - y^2 \). Here, whenever \(x\) and \(y\) sum to more than zero, \(z\) decreases, creating a "bowl" shape.

• The negative signs for \(x^2\) and \(y^2\) indicate the downward direction.
• \(4\) at the vertex signifies its peak height along the \(z\)-axis.

Recognizing how this equation translates into a three-dimensional shape helps in visualizing the solid region.

Volume Calculation

Calculating the volume of our solid involves evaluating the double integral that has been set up. After determining the function and its limits, compute: \[ V = \int_{0}^{1} \int_{-1}^{1}\left(4-x^{2}-y^{2}\right) \, dx \, dy. \] First, perform the integration with respect to \(x\), treating \(y\) as a constant.
This simplifies the expression to a single integral with respect to \(y\).
Next, integrate with respect to \(y\) and evaluate using the limits:

• When integrated over \(x\), terms simplify based on symmetry; terms involving odd powers of \(x\) vanish.
• The individual integrals result in evaluating polynomials like \(8y - \frac{8y^3}{3}\).

Executing these steps results in a final volume of \(\frac{16}{3}\) cubic units. Calculus allows us here to calculate the finite volume under a curved surface, something not as easily done through basic geometry.

Visualizing Solids

Visualizing the described solid can significantly aid comprehension of double integrals and the region they're applied to. The solid in question is a piece of a downward-opening paraboloid. It hovers above the \(xy\)-plane within the constraints given by the integration limits.
In a 3D space, it looks like an upside-down bowl sliced out over a rectangular section on the plane of \(x\) and \(y\).

• The integration borders define a rectangular base in the \(xy\)-plane, meaning \(0 \leq y \leq 1\) and \(-1 \leq x \leq 1\).
• The paraboloid’s full height at the vertex is 4 units, diminishing as \(x\) or \(y\) increase.

Visual tools and graphical software can help sketch this solid, underscoring the practical applicability of the math involved. Recognizing these shapes and their volumetric characteristics enhances spatial understanding of complex three-dimensional regions.