MHB 15.2.30 Where R is the is the region in Q1 & Q4

[karush]

$\textsf{ Evaluate the following integral }$
$\textsf{where $R$ is the is the region in $Q1$ & $Q4$}$
$\textsf{bounded by the semicircle of radius 2 centered at (0,0).}\\$

\begin{align*}\displaystyle
I&=\iint\limits_{R} x^4 y \, dA
\end{align*}
how can this produce a semicircle?

 

[Opalg]

Re: 15.2.30 where R is the is the region in Q1 & Q4

$\textsf{ evaluate the following integral }$
$\textsf{where $r$ is the is the region in $q1$ & $q4$}$
$\textsf{bounded by the semicircle of radius 2 centered at (0,0).}\\$

\begin{align*}\displaystyle
i&=\iint\limits_{r} x^4 y \, da
\end{align*}
how can this produce a semicircle?

I guess the quadrants are numbered in an anticlockwise order, starting with the positive quadrant. So the semicircle would be the half of the circle that lies to the right of the $y$-axis.

    |    
 q2 | q1 
____|____
    |    
 q3 | q4 
    |

 

[karush]

Re: 15.2.30 where R is the is the region in Q1 & Q4

​ok I don' t see where a half circle comes from

 

[Opalg]

Re: 15.2.30 where R is the is the region in Q1 & Q4

​ok I don' t see where a half circle comes from

The circle is the one given in the question, the circle of radius 2 centred at the origin. So you want the right-hand half of that circle. The region $R$ is bounded by that semicircle together with the part of the $y$-axis between $(0,-2)$ and $(0,2)$.

 

[karush]

Re: 15.2.30 where R is the is the region in Q1 & Q4

yes but what is $x^4y$ ?

 

[Opalg]

Re: 15.2.30 where R is the is the region in Q1 & Q4

yes but what is $x^4y$ ?

That is the function that you want to integrate over the region $R$. You will want to use polar coordinates, so $x^4y$ will become $r^5\cos^4\theta\sin\theta$. Form the double integral of that function over $R$.

 

[karush]

Re: 15.2.30 where R is the is the region in Q1 & Q4

That is the function that you want to integrate over the region $R$. You will want to use polar coordinates, so $x^4y$ will become $r^5\cos^4\theta\sin\theta$. Form the double integral of that function over $R$.

OK I'm stabbing in the dark on this one

$\displaystyle\int_{0}^{1} \int_{2\pi/3}^{\pi/2} r^5\cos^4\theta\sin\theta \,dx\,dy$

 

[Opalg]

Re: 15.2.30 where R is the is the region in Q1 & Q4

OK I'm stabbing in the dark on this one

$\displaystyle\int_{0}^{1} \int_{2\pi/3}^{\pi/2} r^5\cos^4\theta\sin\theta \,dx\,dy$

Sorry to have to say it, but just about everything is wrong there.

The region $R$ is the right-hand half of a disc of radius 2 centred at the origin. In terms of polar coordinates, $R$ consists of the points whose polar coordinates $[r,\theta]$ satisfy $0\leqslant r \leqslant 2$ and $-\pi/2 \leqslant\theta\leqslant \pi/2.$ That gives you the limits for the integral, namely $$\int_0^2 \!\!\!\ldots\, dr$$ for the $r$-integral and $$\int_{-\pi/2}^{\pi/2}\!\!\!\ldots\,d\theta$$ for the $\theta$-integral.

Finally, when you convert from Cartesian coordinates to polar coordinates in a double integral, $dx\,dy$ becomes $r\,dr\,d\theta$.

So your integral should be $$\int_0^2\int_{-\pi/2}^{\pi/2} r^6\cos^4\theta\sin\theta \,dr\,d\theta.$$

 

[karush]

Re: 15.2.30 where R is the is the region in Q1 & Q4

So your integral should be $$\int_0^2\int_{-\pi/2}^{\pi/2} r^6\cos^4\theta\sin\theta \,dr\,d\theta.$$

do you mean $r^5$ ?

 

[Opalg]

Re: 15.2.30 where R is the is the region in Q1 & Q4

do you mean $r^5$ ?

No, it's $r^6$. Four of the six come from $x^4 = r^4\cos^4\theta$, and then there is one each from $y = r\sin\theta$ and $dx\,dy = r\,dr\,d\theta.$

 

[karush]

Re: 15.2.30 where R is the is the region in Q1 & Q4

ok, I'm still kinda foggy on this
but similar problems will be soon