# 232.15.1.33 Evaluate the Area of the Region

• MHB
• karush
In summary, the problem asks for the area of a region given as (x, y) such that $0\le x\le 8$ and $0\le y\le ln(4). karush Gold Member MHB$\tiny{232.15.1.33}\\Evaluate the Area of the Region $$f(x,y)=3e^{-y}; R=\biggr[(x,y):0 \le x \le 8, 0 \le y \le \ln 4\biggr]$$ \begin{align*}\displaystyle I&=\iint\limits_{R} f(x,y) \quad dx \, dy \\ &=\int_{0}^{8}\int_{0}^{\ln 4} 3e^{-y} \quad dx \, dy \\ \end{align*} ok just seeing if this is setup ok before proceed. since there is no x in this Last edited: Re: 232.15.1.33 Evaluate the Region I think your integral limits are off. The inner integral is with respect tox$. Re: 232.15.1.33 Evaluate the Region Ackbach said: I think your integral limits are off. The inner integral is with respect to$x. ok so I guess we can just switch x to y like thus: \begin{align*}\displaystyle &=\int_{0}^{8}\int_{0}^{\ln 4} 3e^{-y} \quad dy \, dx \\ \end{align*} Re: 232.15.1.33 Evaluate the Region karush said: ok so I guess we can just switchx$to$ylike thus: \begin{align*} &=\int_{0}^{8}\int_{0}^{\ln 4} 3e^{-y} \, dy \, dx \\ \end{align*} Yes, but only because your region is rectangular, andy$and$x$are independent on the boundary of your region. If you had, say, a triangular region, you'd have to think carefully about it. For example, suppose you had $$\int_0^1 \int_0^x f(x,y) \, dy \, dx.$$ You can draw this triangular region out - it's a right triangle with vertices at$(0,0), \; (1,0),\; (0,1). Now suppose you wanted to interchange the order of integration. How would you do that? Like this: $$\int_0^1 \int_y^1 f(x,y) \, dx \, dy.$$ Well, that doesn't look anything like the previous one! Why is that? Because the inner integral has to be viewed as going between two functions as limits, not two numbers. Does that help? \begin{align*}\displaystyle I&=\iint\limits_{R} f(x,y) \quad dx \, dy \\ &=\int_{0}^{8}\int_{0}^{\ln 4} 3e^{-y} \quad dx \, dy \\ &=\int_{0}^{8}\biggr[-3e^{-y}\biggr]_0^{\ln{ 4}}\quad dy=\int_{0}^{8}\frac{9}{4}\quad dy\\ &=9\biggr[\frac{y}{4}\biggr]_0^8=9\cdot (2-0)\\ &=\color{red}{\large{18}} \end{align*}hopefully any suggest karush said: \begin{align*}\displaystyle I&=\iint\limits_{R} f(x,y) \quad dx \, dy \\ &=\int_{0}^{8}\int_{0}^{\ln 4} 3e^{-y} \quad dx \, dy \\ &=\int_{0}^{8}\biggr[-3e^{-y}\biggr]_0^{\ln{ 4}}\quad dy=\int_{0}^{8}\frac{9}{4}\quad dy\\ &=9\biggr[\frac{y}{4}\biggr]_0^8=9\cdot (2-0)\\ &=\color{red}{\large{18}} \end{align*}hopefully any suggest Check your answer on Wolfram|Alpha or Wolfram Development Platform with this command: Code: Integrate[Integrate[3*Exp[-y],{x,0,Log[4]}],{y,0,8}] What, exactly, was the wording of the problem? You titled this, and the first line of your post is "Find the area of the region" but then introduce a function f(x, y). Is the problem to find the area or is it to integratef(x, y)= 3e^{-y}$over that region? The region is given as the set of all (x, y) such that$0\le x\le 8$and$0\le y\le ln(4)$. That region is a rectangle. The area of that rectangle is, of course, 8ln(4). The integral of f(x,y) over that region can be done as$\int_0^8 \int_0^{ln(4)} 3e^{-y} dy dx$or$\int_0^{ln(4)}\int_0^8 3e^{-y} dx dy$or$\left(\int_0^8 dx\right)\left(\int_0^{ln(4)} 3e^{-y}dy\right)$. If you were concerned, in doing$\int_0^{ln(4)}\int_0^8 3e^{-y} dxdy$that "there is no x in that", you treat it as the integral of a constant.$\int_0^8 3e^{-y} dx= 3e^{-y}\left[x\right]_0^8= 8(3e^{-y})$. Then$\int_0^{ln(4)} 8(3e^{-y}) dy= 24 \int_0^{ln(4)} e^{-y} dy= 24\left[-e^{-y}\right]_0^{ln(4)}= 24(-e^{-ln(4)}+ e^0)= 24(1- \frac{1}{4}= 24(\frac{3}{4})= 18\$ as you got.

Last edited:
Ackbach said:
Check your answer on Wolfram|Alpha or Wolfram Development Platform with this command:

Code:
Integrate[Integrate[3*Exp[-y],{x,0,Log[4]}],{y,0,8}]

well how come its not 18

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## 1. What does the IP address 232.15.1.33 refer to?

The IP address 232.15.1.33 is a numerical label assigned to a device connected to a network, specifically the Internet Protocol (IP) network. It is used to identify and communicate with the device on the network.

## 2. What does it mean to evaluate the area of a region?

Evaluating the area of a region means finding the measurement of the flat surface inside the boundaries of a shape or region. This can be done using various mathematical methods and formulas, depending on the shape and complexity of the region.

## 3. How is the area of a region calculated?

The method of calculating the area of a region depends on the shape of the region. For example, the area of a rectangle is calculated by multiplying the length by the width, while the area of a circle is calculated by multiplying the square of the radius by pi (3.14).

## 4. What factors can affect the accuracy of evaluating the area of a region?

The accuracy of evaluating the area of a region can be affected by factors such as the precision of measurements, the complexity of the shape, and the accuracy of the mathematical methods used. Human error and limitations in technology can also contribute to a margin of error in the calculation.

## 5. Why is evaluating the area of a region important in science?

Evaluating the area of a region is important in science because it allows scientists to accurately measure and describe the physical properties of objects and the natural world. It is used in various fields of science, such as physics, biology, and geography, to analyze and understand the characteristics and behavior of different systems and phenomena.

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