MHB 232.15.1.33 Evaluate the Area of the Region

[karush]

$\tiny{232.15.1.33}\\$
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

 

[Ackbach]

Re: 232.15.1.33 Evaluate the Region

I think your integral limits are off. The inner integral is with respect to $x$.

 

[karush]

Re: 232.15.1.33 Evaluate the Region

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*}

 

[Ackbach]

Re: 232.15.1.33 Evaluate the Region

ok so I guess we can just switch $x$ to $y$ like thus:
\begin{align*}
&=\int_{0}^{8}\int_{0}^{\ln 4} 3e^{-y} \, dy \, dx \\
\end{align*}

Yes, but only because your region is rectangular, and $y$ 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?

 

[karush]

\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

 

[Ackbach]

\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:

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

 

[HOI]

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 integrate $f(x, y)= 3e^{-y}$ 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.

 

[karush]

Check your answer on Wolfram|Alpha or Wolfram Development Platform with this command:

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

ok well I already had it in latex...
well how come its not 18

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