Problem 46 - Thomas Calculus

Question

Find the volume of the solid cut from the first octant by the surface \(z=4-x^{2}-y\)

Step-by-Step Solution

Answer

The volume is \(\frac{208}{15}\).

Step 1: Understand the Problem

The first octant means the region where all coordinates \((x, y, z)\) are non-negative. The surface given is defined by the equation \(z = 4 - x^2 - y\). We need to determine the volume underneath this surface within the bounds imposed by the non-negative axes in the first octant.

Step 2: Determine the Boundaries

We need to find where the surface intersects the xy-plane, which happens where \(z = 0\). Setting \(z = 0\) in the equation \(z = 4 - x^2 - y\) gives \(4 - x^2 - y = 0\), or equivalently, \(y = 4 - x^2\). This equation bounds the region in the xy-plane we are considering.

Step 3: Set Up the Double Integral

We're aiming to find the volume under the surface in the region determined in Step 2. The volume can be represented by the integral \(V = \int \int_R (4 - x^2 - y) \, dx \, dy\), where \(R\) is the region in the xy-plane bounded by \(y = 4 - x^2\), \(y = 0\), and \(x = 0\).

Step 4: Determine Limits of Integration

From the intersection line \(y = 4 - x^2\), the limits for \(y\) vary from 0 to \(4 - x^2\). For \(x\), because \(y = 4 - x^2\) meets the x-axis where \(4 - x^2 = 0\) or \(x^2 = 4\), \(x\) will vary from 0 to 2. Therefore, the limits are \(0 \leq x \leq 2\) and \(0 \leq y \leq 4 - x^2\).

Step 5: Compute the Inner Integral

Compute the integral with respect to \(y\): \[ \int_0^{4 - x^2} (4 - x^2 - y) \, dy = \left[ 4y - x^2y - \frac{y^2}{2} \right]_0^{4-x^2} \]. Evaluate this expression to get \[ 16 - 8x^2 + \frac{x^4}{2}\].

Step 6: Compute the Outer Integral

Now integrate the result from the inner integral with respect to \(x\): \[ \int_0^2 (16 - 8x^2 + \frac{x^4}{2}) \, dx \]. This evaluates as follows:- \(\int_0^2 16 \, dx = 16x \big|_0^2 = 32\)- \(\int_0^2 8x^2 \, dx = \frac{8x^3}{3} \big|_0^2 = \frac{64}{3}\)- \(\int_0^2 \frac{x^4}{2} \, dx = \frac{x^5}{10} \big|_0^2 = \frac{32}{10}\)Thus, the volume is \(32 - \frac{64}{3} + \frac{16}{5}\).

Step 7: Calculate Final Volume

Calculate the final volume: - Convert the fractions \(32 - \frac{64}{3} + \frac{16}{5}\) to a common denominator and evaluate. - \(32 = \frac{480}{15}\), \(\frac{64}{3} = \frac{320}{15}\), \(\frac{16}{5} = \frac{48}{15}\).Now combine and simplify: \(\frac{480}{15} - \frac{320}{15} + \frac{48}{15} = \frac{208}{15}\).

Key Concepts

Double Integral

Double integrals are a powerful tool used to calculate the volume under a surface over a specific region in the plane. The basic idea is similar to that of a single integral, which measures the area under a curve. A double integral extends this concept to two dimensions, allowing us to determine the volume under a surface defined by a function of two variables.
The double integral is represented as \(\int \int_R f(x, y) \, dx \, dy\), where \(R\) is the region over which we are integrating, and \(f(x, y)\) is the function describing the height of the surface at any point \((x, y)\) within \(R\).
To compute a double integral, we typically perform two iterations of integration: one with respect to \(y\) (the inner integral) and another with respect to \(x\) (the outer integral). In this way, we cover all points in the region \(R\) and sum up the contributions to the volume.

First Octant

The first octant in three-dimensional space is an important concept when considering volumes under surfaces. It refers to the section of space where all coordinate values are non-negative, that is \((x, y, z) \geq 0\).
Visualize the three-dimensional Cartesian coordinate system. The first octant is akin to the top-right corner of this space where the positive x-axis, y-axis, and z-axis all meet. In exercises like the one we're discussing, the first octant forms a natural boundary, ensuring that all coordinates are either on or above the plane created by the origin.
In our specific problem, we consider a surface that's situated in the first octant, meaning that the limits of integration must respect this boundary condition. Only positive coordinates matter to us, which influences both the region \(R\) and the bounds of our double integral.

Limits of Integration

Setting the correct limits of integration is crucial for solving any problem involving double integrals. These limits define the region \(R\) over which we integrate and are typically represented as \(a \leq x \leq b\) and \(g(x) \leq y \leq h(x)\).
In the given problem, we first replace \(z\) with zero to find where the surface intersects the xy-plane. This gives us the equation \(y = 4 - x^2\), defining the upper boundary for \(y\). Meanwhile, both \(x\) and \(y\) are constrained to be non-negative because we are working within the first octant.
The determined limits are therefore \(0 \leq x \leq 2\) and \(0 \leq y \leq 4 - x^2\). These bounds help ensure that we integrate over the correct region to obtain the volume of interest.

Intersection with Planes

Understanding the intersection of surfaces with planes is vital for determining regions of integration. The intersection point can indicate where a 3D surface touches a particular plane, such as the xy-plane, which simplifies the evaluation of integrals.
For the problem at hand, we consider the surface defined by \(z = 4 - x^2 - y\). To find the intersection with the xy-plane, we set \(z = 0\). Solving the equation \(4 - x^2 - y = 0\) reveals that the surface intersects with the plane along the curve \(y = 4 - x^2\).
This intersection defines the region in the xy-plane over which our double integral is calculated, ensuring that we only integrate over points that lie on or below the surface. By understanding such intersections, we can correctly establish the geometric region required for the integration and hence accurately determine the volume of the solid in question.