Question

Compute the volume of the following solids. Square column The column with a square base \(R=\\{(x, y):|x| \leq 1,|y| \leq 1\\}\) cut by the plane \(z=4-x-y\)

Short answer

Question: Determine the volume of the solid formed by the square column cut by the plane z=4-x-y. Answer: The volume of the given solid is 1 cubic unit.

Step-by-step solution

Set up the integral bounds

Since our base is a square with vertices at \((-1,-1), (1,-1), (1,1),\) and \((-1,1),\) the integral bounds will vary between -1 and 1 for both x and y.

Set up the integrand function

The given plane is \(z=4-x-y\). This will be the height function that will be integrated over the given region.

Set up the double integral

Thus, the volume of the solid can be written as a double integral using the function \(z=4-x-y\) over the given region: $$V = \int_{-1}^1 \int_{-1}^1 (4-x-y) \, dy \, dx$$.

Evaluate the inner integral

Now, we evaluate the inner integral: $$\int_{-1}^1 (4-x-y) \, dy = [(4-x)y - \frac{1}{2}y^2]_{-1}^1$$.

Simplify the expression

Simplifying the expression inside the brackets gives us: $$((4-x)(1)-\frac{1}{2}(1)^2) - ((4-x)(-1)-\frac{1}{2}(-1)^2) = 4-2x - 3x = 1 - x$$.

Evaluate the outer integral

Now that we have simplified the expression, we can evaluate the outer integral: $$V = \int_{-1}^1 (1 - x) \, dx$$.

Integrate and find the volume

Finally, we integrate the function and obtain the volume of the solid: $$V = [x - \frac{1}{2}x^2]_{-1}^1 = (1 - \frac{1}{2}) - (-1 - \frac{1}{2}) = 1 - \frac{1}{2} + 1 - \frac{1}{2} = 1$$. Thus, the volume of the given solid is 1 cubic unit.

Key concepts

Double Integral

The concept of a double integral may sound complex, but it's fundamentally about calculating an area under a surface within two dimensions. In our example, it helps us find the volume under a plane above a specific region on the xy-plane. Instead of merely integrating over one line, we integrate over an entire area.
If you visualize a 3D shape, a double integral allows you to 'stack' infinitesimally small pieces over a region, which when accumulated gives the entire volume. The process involves proceeding with integration in two steps, typically with respect to two variables, such as x and y, in our given area.
For example, we used:

• First integrate with respect to y, keeping x constant. Then, integrate the resulting function with respect to x.

Thus, double integrals offer a method to collect parts of a volume that lie over a defined area, essentially "summing" over two dimensions.

Integral Bounds

Integral bounds define the limits over which you take the integral. They confine the region within which the function is integrated. In our problem, the base of the solid is a square, specified by the bounds of x and y values.
Each range from

• -1 to 1 for x
• -1 to 1 for y

These limitations tell us the region of interest in the xy-plane. Ultimately, integral bounds delineate the area above which the double integral captures the volume, ensuring the calculations pertain precisely to the region under the plane near the solid's base.

Evaluating Integrals

Evaluating integrals is the culmination of setting up a problem that requires integration. Here, it involves two rounds of integration: first the inner, and then the outer integral. This sequential approach is crucial for deriving the exact volume.
Initially, you tackle the integral with respect to one variable (y, in this case), treating the other variable (x) as a constant. This step simplifies the function.

• For the inner integral: egin{equation} rac{d}{dy}igg((4-x-y)igg) ext{ evaluated from } -1 ext{ to } 1 igg] = 1-x. igg. igg) igg) igg) < bigg]<]+tabobile the functions and determine that . li
• For the outer integral: egin{equation} rac{d}{dx}igg((1-x)igg) ext{ evaluated from } -1 ext{ to } 1 igg] = 2]. igg]. igg) igg] igg) ](abmobil itime iterms x a function validates the problem of finding<) =hangewissing clear story out of integrals that let you find Through incrementally following these steps, the double integral simplifies down to determining the precise volume of the specified solid, arriving at one definitive cubic unit.