Question

Find the volume of the following solid regions. The solid \(S\) between the surfaces \(z=e^{x-y}\) and \(z=-e^{x-y},\) where \(S\) intersects the \(x y\) -plane in the region \(R=\\{(x, y): 0 \leq x \leq y\) \(0 \leq y \leq 1\\}\)

Short answer

Question: Find the volume of the solid region S enclosed between the surfaces z = e^(x-y) and z = -e^(x-y), with a projection region R on the xy-plane given as a triangle with vertices (0,0), (0,1), and (1,1). Answer: The volume of the solid region S is 4 - 2e^(-1).

Step-by-step solution

Visualize the problem

First, it's helpful to create a mental (or actual) image of the region \(R\) on the \(xy\)-plane, where \(S\) intersects it. This region is a right triangular region, bounded by the lines \(y = x, y = 1\) and the \(y\)-axis.

Express the volume integral

Since we want to find the volume of the solid region enclosed between the two surfaces, we can express the volume integral as: $$V = \iint_R \Big(e^{x-y} - (- e^{x-y}) \Big) \, dA$$

Convert to an iterated integral

To make it an iterated integral, we need to figure out the limits for \(x\) and \(y\) based on our region \(R\). As \(0 \leq x \leq y\) and \(0 \leq y \leq 1\), we can write $$V = \int_{0}^{1} \int_{0}^{y} \Big(e^{x-y} - (- e^{x-y}) \Big) \, dx\,dy$$

Evaluate the inner integral

Now, we need to evaluate the inner integral by treating the outer variable \(y\) as a constant: $$\int_{0}^{y} \Big(e^{x-y} - (- e^{x-y}) \Big) \, dx = \left[ e^{x-y} - (-e^{x-y}) \right]_{0}^{y} = 2 \Big(e^{y - y} - e^{0 - y} \Big) = 2 \Big(e^0 - e^{-y} \Big)$$

Evaluate the outer integral

Now we have a single integral left to evaluate: $$V = 2 \int_{0}^{1} \Big(e^0 - e^{-y} \Big) \, dy$$ Evalute this integral: $$V = 2 \left [\int_{0}^{1} 1 \, dy - \int_{0}^{1} e^{-y} \, dy \right] = 2 \left[ y - (-e^{-y}) \right]_{0}^{1} = 2 \left[ \left(1 - (-e^{-1}) \right) - \left( 0 - (-e^0) \right) \right]$$ Finally, simplify our result: $$V = 2 \left(1 - e^{-1} + e^0 \right) = 2 (2 - e^{-1}) = 4 - 2 e^{-1}$$ Therefore, the volume of the solid region \(S\) is: $$V = 4 - 2 e^{-1}$$

Key concepts

Volume Integral

To find the volume of a solid region bound by surfaces, you use the concept of a volume integral. Essentially, it's an extension of the familiar area integral used for two-dimensional regions, into three dimensions. The solid volume between two surfaces can be calculated by integrating the difference between the upper and lower surface functions over the specified region in the plane. In our problem, the surfaces are given by the functions:

• Upper surface: \( z = e^{x-y} \)
• Lower surface: \( z = -e^{x-y} \)

The volume integral is written as: \[ V = \iint_R \Big(e^{x-y} - (-e^{x-y}) \Big) \, dA \] Here, \( R \) is the projection of the solid onto the \(xy\)-plane, a triangular region in this case. Understanding this integral involves seeing it as a summation: each tiny rectangular prism within the solid adds a little to the total volume, and as such, we integrate over the whole region \( R \).

Iterated Integral

An iterated integral is a technique to evaluate double integrals by iteratively integrating one variable at a time. This approach simplifies calculating the volume of a solid by breaking down a double integral into two separate, successive integrations. Our task splits into nested integrals, representing the area in which the solid is located: \[ V = \int_{0}^{1} \int_{0}^{y} \Big(e^{x-y} - (-e^{x-y}) \Big) \, dx \, dy \] The process involves evaluating the inner integral with respect to one variable (here, \(x\)) while treating the other variable (\(y\)) as a constant. Post this, the outer integral is evaluated. Using iterated integrals helps manage complex region calculations by addressing dimensions step by step, reducing overall computational complexity.

Exponential Functions

Exponential functions frequently appear in calculus problems due to their unique characteristics, particularly the rate of change. The functions in the current problem, \( e^{x-y} \) and \( -e^{x-y} \), illustrate their rapid increase or decrease as the variables change. A key property of exponential functions is that their derivative with respect to each variable simply consists of the same exponential function, allowing them to "smoothly" expand across the axes without any sharp changes or discontinuities. Moreover, these functions often appear in integrals due to their ease of differentiation, making them highly manageable for calculation. For example, when the function \((e^{-y})\) was integrated, its antiderivative quickly led to a solvable expression, highlighting the convenience of exponential properties.

Triangular Region

Any structural visualization of a problem greatly aids in comprehending the underlying geometric boundaries. Here, the triangular region \( R \) in the \(xy\)-plane is defined as \( \{(x, y): 0 \leq x \leq y, \ 0 \leq y \leq 1\} \). This is a right triangle bounded by the lines \( y = x \), \( y = 1 \), and the \( y \)-axis. Visualizing this region geometrically aids in correctly setting the limits for the iterated integral. Understanding how these boundaries dictate which "slices" of the solid have to be computed ensures the area over which the computation truly runs fits in line with the specified conditions. By working with triangles, integration boundaries can be adjusted more readily compared to complex shapes, simplifying calculations while maintaining accuracy.