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?