Problem of double integrals in polar coordinates

[Seimuna]

2. $\displaystyle \int \int 2xy dA$ where R is bounded by $\displaystyle y=4-x^{2}, \;\ and \;\ y=0$

so, does the substitution go this way?

$\displaystyle \int_{0}^{2\pi}\int_{0}^{4-x^{2}}[2(r sin{\theta})(r cos{\theta})] r dr d{\theta}$ ???

You forgot to change the $\displaystyle 4-x^{2}$ to polar form. It is still in rectangular format.

$\displaystyle 4-r^{2}cos^{2}{\theta}$

 

[galactus]

1.$\displaystyle \sqrt{x^{2}+ y^{2} + 1} dA$ where R is the disk $\displaystyle x^{2}+y^{2}\leq 16$

so, is the solution lik this?
$\displaystyle \int_{0}^{2\pi}\int_{0}^{4}r\sqrt{r^{2}+1}drd{\theta}$
$\displaystyle \int_{0}^{2\pi} (1/3) [17^{\frac{3}{2}}-1] d{\theta}$
$\displaystyle = \frac{2\pi}{3} [17^{\frac{3}{2}}-1]\approx 144.7076$

yes, you got it. Good work .

I changed your work to LaTex format so it would be easier to read.

BTW, for your benefit, that Greek letter is spelled 'theta', not tita.

In rectangular, the equivalent is $\displaystyle \int_{-4}^{4}\int_{-\sqrt{16-x^{2}}}^{\sqrt{16-x^{2}}}\sqrt{x^{2}+y^{2}+1}dydx$

But it is harder to deal with than the polar. That is the reason we change coordinate systems.

 

[Seimuna]

2. $\displaystyle \int \int 2xy dA$ where R is bounded by $\displaystyle y=4-x^{2}, \;\ and \;\ y=0$

so, does the substitution go this way?

$\displaystyle \int_{0}^{2\pi}\int_{0}^{4-x^{2}}[2(r sin{\theta})(r cos{\theta})] r dr d{\theta}$ ???

You forgot to change the $\displaystyle 4-x^{2}$ to polar form. It is still in rectangular format.

$\displaystyle 4-r^{2}cos^{2}{\theta}$

wow!!! i make it... im so happy... thanks!!!
erm...actually, i feel confuse of how to define the limit for "theta" and "r"... i wonder when the lower limit shud be a zero or a value... im jus assuming for all theta - the upper limit is 2pi and lower limit is zero and for "r" - the upper limit - get from the x^2+y^2 equation and the lower limit, normally i ll assume as zero as well...

erm...another question...how to type the question into LaTex form?

 

[galactus]

Click on 'quote' at the upper right corner of my post to see the code I used.

 

[Seimuna]

[quote:1j6uzg4h]$\displaystyle \int \int 2xy dA$ where R is bounded by $\displaystyle y=4-x^{2}, \;\ and \;\ y=0$

so, does the substitution go this way?

$\displaystyle \int_{0}^{2\pi}\int_{0}^{4-x^{2}}[2(r sin{\theta})(r cos{\theta})] r dr d{\theta}$ ???

You forgot to change the $\displaystyle 4-x^{2}$ to polar form. It is still in rectangular format.

$\displaystyle 4-r^{2}cos^{2}{\theta}$[/quote:1j6uzg4h]

im so sorry...
if substitute $\displaystyle 4-r^{2}cos^{2}{\theta}$ to the upper limit, then, when substitute the limit to the r, it ll stil having r in the equation... wat shud i do?