Question

Find the volume between the planes $z=2x+3y+6$ and $z=2x+7y+8,$ and over the triangle with vertices $(0,0),(3,0),$ and (2,1).

Short answer

The volume is 4.5 cubic units.

Step-by-step solution

Define the Region

Identify the triangular region over which the volume needs to be calculated. The vertices are given as (0,0), (3,0), and (2,1). This defines a triangle in the XY-plane.

Set Up the Volume Integral

The volume between two surfaces can be found using a double integral over the region R, with the integrand being the difference between the two surfaces: $$V={\iint }_R(f_2(x,y)-f_1(x,y))dA$$ where $$f_1(x,y)=2x+3y+6$$ and $$f_2(x,y)=2x+7y+8$$

Calculate the Difference Between the Surfaces

Calculate the difference between the two plane equations: $$f_2(x,y)-f_1(x,y)=(2x+7y+8)-(2x+3y+6)=4y+2$$

Set Up the Limits of Integration

Determine the limits of integration for the triangular region. For this triangle:- The limits for x are from 0 to 3.- For a given x, y ranges from 0 to the line equation from (3,0) to (2,1), which is y=1-$\frac{1}{2}$x. Thus, the integral is set up as: $${\int }_0^3{\int }_0^{1-\frac{1}{2}x}(4y+2)dydx$$

Integrate with Respect to y

Evaluate the inner integral with respect to y: $${\int }_0^{1-\frac{1}{2}x}(4y+2)dy={[2y^2+2y]}_0^{1-\frac{1}{2}x}=2(1-\frac{1}{2}x{)}^2+2(1-\frac{1}{2}x)$$

Simplify the Result

Simplify the expression from the previous step: $$2(1-\frac{1}{2}x{)}^2+2(1-\frac{1}{2}x)=2(1-x+\frac{1}{4}{x}^2)+2-x=2-2x+\frac{1}{2}{x}^2+2-x=4-3x+\frac{1}{2}{x}^2$$

Integrate with Respect to x

Evaluate the outer integral: $${\int }_0^3(4-3x+\frac{1}{2}{x}^2)dx={[4x-\frac{3}{2}{x}^2+\frac{1}{6}{x}^3]}_0^3=12-\frac{27}{2}+\frac{9}{2}=12-9+1.5=4.5$$

Key concepts

double integral

A double integral allows us to compute the volume between two surfaces over a specific region in the XY-plane. It's an extension of the concept of an integral from single-variable calculus to two variables.
In this problem, we use a double integral to find the volume between two planes over a triangular region by integrating the difference of their equations. The double integral is represented as:
$$V={\iint }_R(f_2(x,y)-f_1(x,y))dA$$
This formula calculates the combined areas of small vertical slices between the planes, summing up for the entire region.
To proceed, we must define the region, the difference between the surfaces, and the limits of integration for accurate volume determination.

limits of integration

Setting up the correct limits of integration is essential to accurately evaluating a double integral. The limits define the boundaries within which we integrate. In our exercise, the region is a triangle with vertices $0,0$, $3,0$, and $2,1$.
We calculate limits as follows:

• The limits for $x$ are straightforward: $$0\le x\le 3$$
• For $y$, dependent on $x$, we find the range using the line equations. Here, $$y=1-\frac{1}{2}x$$
  Thus, $$0\le y\le 1-\frac{1}{2}x$$
  Plugging these limits into our double integral setup, we have:
  $${\int }_0^3{\int }_0^{1-\frac{1}{2}x}(4y+2)dydx$$

Accurate limits ensure the entire specified region is covered and nothing outside it is considered.

calculating volume between surfaces

To calculate the volume between two surfaces, we subtract the lower surface equation from the upper one and then integrate this difference over the given region.
In the given problem, the planes are represented by:

• $$f_1(x,y)=2x+3y+6$$
• \_$$f_2(x,y)=2x+7y+8$$

Thus, their difference is:
$$f_2(x,y)-f_1(x,y)=(2x+7y+8)-(2x+3y+6)=4y+2$$
This difference, $$(4y+2)$$, becomes our integrand.
We then proceed step-by-step:

• Calculate the inner integral with respect to y.
• Simplify the result.
• Evaluate the outer integral with respect to x.

The final outcome gives us the volume between the specified planes over the triangular region.

triangular region in XY-plane

Understanding the triangular region in the XY-plane is key in setting up our integral. The given vertices are:

• $(0,0)$
• $(3,0)$
• $(2,1)$

These points define the area where we perform integration.
First, we find the boundary equations of the triangle:

• The line from $(0,0)$ to $(2,1)$ can be written as $$y=\frac{1}{2}x$$
• The line from $(3,0)$ to $2,1)$ is more relevant, given by $$y=1-\frac{1}{2}x$$
  These equations help us set our integration limits.
  Visualizing and understanding these lines is crucial for correctly setting up a double integral over the specified regions.