Problem 62 - 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 approximately 8.535 cubic units.

Step 1: Identify the region in the xy-plane

For the first octant, we have the constraints that all coordinates are non-negative: \\( x \geq 0 \), \( y \geq 0 \), and \( z \geq 0 \). From the surface equation \( z = 4 - x^2 - y \), for \( z \geq 0 \), it follows that \( 4 - x^2 - y \geq 0 \), hence the inequality \( y \leq 4 - x^2 \). This describes the region in the xy-plane over which we will integrate to find the volume.

Step 2: Set up the double integral for volume

The volume under the surface \( z = 4 - x^2 - y \) within the specified region can be found using the double integral:\[V = \int_{0}^{2} \int_{0}^{4-x^2} (4 - x^2 - y) \, dy \, dx\]The outer integral bounds correspond to the range of \( x \), from 0 to 2 determined by \( 4 - x^2 = 0 \), and the inner integral bounds represent \( y \)'s limits from 0 to \( 4 - x^2 \).

Step 3: Integrate with respect to y

Perform the integration with respect to \( y \): \[\int_{0}^{4-x^2} (4 - x^2 - y) \, dy = \left[ (4 - x^2)y - \frac{y^2}{2} \right]_{0}^{4-x^2}\]This simplifies to:\[(4-x^2)(4-x^2) - \frac{(4-x^2)^2}{2}\]This reduces to simplification of the expression.

Step 4: Simplify the expression

Simplify the expression obtained from the previous step:\[(4-x^2)(4-x^2) - \frac{(4-x^2)^2}{2} = \frac{(4-x^2)^2}{2}\]This represents the area for slice at \(x\).

Step 5: Integrate with respect to x

Now integrate the simplified expression with respect to \( x \):\[V = \frac{1}{2} \int_{0}^{2} (4 - x^2)^2 \, dx\]Expand and integrate the polynomial:\[V = \frac{1}{2} \int_{0}^{2} (16 - 8x^2 + x^4) \, dx = \frac{1}{2} \left[ 16x - \frac{8x^3}{3} + \frac{x^5}{5} \right]_{0}^{2}\]

Step 6: Substitute and compute the definite integral

Substitute the bounds into the integrated function:\[= \frac{1}{2} \left[ (16 \times 2) - \frac{8 \times 2^3}{3} + \frac{2^5}{5} - (0) \right] = \frac{1}{2} (32 - \frac{64}{3} + \frac{32}{5})\]Calculate the numeric value of the expression.

Step 7: Simplify the numeric expression

Simplify the calculated expression to find the volume:\[= \frac{1}{2} \left( 32 - \frac{64}{3} + \frac{32}{5} \right)= \frac{1}{2} \left( 32 - 21.33 + 6.4 \right)= \frac{1}{2} \left( 17.07 \right)= 8.535\]

Step 8: Finalize and review

The computed volume of the solid in the first octant under the surface \( z = 4 - x^2 - y \) is 8.535 cubic units. Ensure that each integration and simplification step was correctly completed to verify the final result.

Key Concepts

Double Integrals

When calculating the volume of a solid bounded by a surface, especially in three dimensions, double integrals are indispensable. They allow us to sum up infinitesimally small pieces of an area in the xy-plane and extend that into the third dimension.
To visualize, think of the xy-plane as a flat sheet of paper. For each tiny square, or differential area, on this paper, we project a height determined by our integrand (in this case, the function \( z = 4 - x^2 - y \)).
This process creates thin slices, or columns, which we add up to find the total volume.

• The function within the integral gives the height of these slices.
• The order and limits of integration dictate the shape and extent of these slices.

Understanding double integrals means grasping how two integrations work together to describe a three-dimensional shape.

Iterated Integrals

Iterated integrals are a specific type of double integral that breaks the process into two distinct steps. This means you first integrate with respect to one variable and then take the result and integrate with respect to the second variable.
In the example exercise, we used iterated integrals to first integrate with respect to \( y \) and then with respect to \( x \).

• The inner integral (first integration) defines a curve depending on the value of the outer variable.
• The outer integral then sums these curves over the specified bounds of the outer variable.

This method is particularly useful when dealing with functions that describe surfaces above a region in the xy-plane. Iterated integrals help conveniently manage the complexity of such multi-dimensional problems.

Integration Bounds

Setting proper integration bounds is crucial for solving problems involving volume using double integrals. These bounds define the region over which we integrate.
For the problem at hand, we need to consider the region in the first octant where the function limits us by the plane \( z = 0 \).

• The outer bounds, for \( x \), were determined from setting the function equal to zero: \( 4 - x^2 = 0 \).
• The inner bounds are determined by the inequality derived from the original function: \( y \leq 4 - x^2 \).

Each integral bound affects the shape of the volume we calculate. Correctly defining these boundaries ensures that the calculated volume matches the actual physical space defined by the problem.

First Octant

The first octant in three-dimensional space is defined where all three variables \( x, y, \) and \( z \) are non-negative. Imagine a corner of a room extending infinitely upwards and to the sides, bounded only by the floor and walls.
This constraint simplifies calculations since we deal solely with positive numbers, often making it easier to define integration limits.

• This way of setting limits confines our calculative exploration to positive regions in space.
• In our solution, \( x \geq0 \), \( y\geq0 \), and \( z\geq0 \).

Understanding the concept of the first octant is essential, as it influences how we visualize the problem and set up our integrals.

Volume Calculation

Calculating volume using integrals involves summing the 'pieces' or 'slices' of volume defined by the region's shape above the integration area.
For the given function \( z = 4 - x^2 - y \), we looked at how to slice the region incrementally, integrating first in the \( y \)-direction and then in the \( x \)-direction. This gives a final numerical value representing the total volume enclosed by the surface down to the region in the xy-plane.

• Each integration step reduces the problem's dimensionality until it's collapsed into the numeric final result.
• The process involves careful computation of expressions derived from the limits and bounds set out initially.

Ultimately, through a calculated series of integral operations, we arrive at the solid's precise volume, demonstrating a powerful application of calculus in understanding three-dimensional spaces.