MHB 15.1.25 Evaluate the following double integral over the region R

[karush]

$\tiny 15.1.25$
$\textsf{Evaluate the following double integral over the region R}\\$
$\textit{note: the R actually is supposed be under both Integrals don't know the LaTEX for it}$
\begin{align*}\displaystyle
\int_R\int&=5(x^5 - y^5)^2 dA\\
R&=[(x,y): 0 \le x \le 1, \, -1 \le y \le -1]
\end{align*}

OK pretty new at this and just did a few previous problems
which were double integrals but different

 

[MarkFL]

As for the $\LaTeX$, look under the "Calculus/Analysis" section of our Quick $\LaTeX$ tool. You will find it, which gives the code:

\iint\limits_{}

into which you can enter "R":

\iint\limits_{R}

to give:

$$\iint\limits_{R}$$

Now, when you sketch the region $R$, what do you find?

 

[karush]

$\tiny 15.1.25$
$\textsf{Evaluate the following double integral over the region R}\\$
\begin{align*}\displaystyle
&\iint\limits_{R}5(x^5 - y^5)^2 dA\\
R&=[(x,y): 0 \le x \le 1, \, -1 \le y \le -1]
\end{align*}

View attachment 7244

I don't know why y goes to -1

 

[MarkFL]

The region $R$ is a rectangle having the vertices:

$$(0,1),\,(1,1),\,(1,-1),\,(0,-1)$$

One way we can compute the volume is:

$$V=\int_{-1}^{1}\int_0^1 5\left(x^5-y^5\right)^2\,dx\,dy=5\int_{-1}^{1}\int_0^1 x^{10}-2x^5y^5+y^{10}\,dx\,dy=5\int_{-1}^{1}\left[\frac{x^{11}}{11}-\frac{x^6y^5}{3}+xy^{10}\right]_0^1\,dy=$$

$$5\int_{-1}^{1} \frac{1}{11}-\frac{y^5}{3}+y^{10}\,dy=5\left[\frac{y}{11}-\frac{y^6}{18}+\frac{y^{11}}{11}\right]_{-1}^{1}=5\left(\left(\frac{1}{11}-\frac{1}{18}+\frac{1}{11}\right)-\left(-\frac{1}{11}-\frac{1}{18}-\frac{1}{11}\right)\right)=\frac{20}{11}$$

 

[karush]

thank you so much for showing all the steps

I was trying to deal with it without squaring which made it a lot harder

was cool to type in the answer and see the happy green pop up from MML

they have examples but it is really complicated. its tons better just to come here

 

[HallsofIvy]

"[math]-1\le y\le -1[/math]" makes no sense. Are you sure it is not supposed to be "[math]-1\le y\le 1[/math]"? As given the "region" of integration is a line segment, not a region at all, and the double integration over that is 0.

 

[MarkFL]

"[math]-1\le y\le -1[/math]" makes no sense. Are you sure it is not supposed to be "[math]-1\le y\le 1[/math]"? As given the "region" of integration is a line segment, not a region at all, and the double integration over that is 0.

Good catch! I didn't even see that...I saw what I expected. :)

 

[karush]

"[math]-1\le y\le -1[/math]" makes no sense. Are you sure it is not supposed to be "[math]-1\le y\le 1[/math]"? As given the "region" of integration is a line segment, not a region at all, and the double integration over that is 0.

it should be $-1 \le y \le 1$
hard to see typo