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eduard
2 years ago
15

∫∫(x+y)dxdy ,d là miền giới hạn bởi x²+y²=1

Mathematics
1 answer:
igor_vitrenko [27]2 years ago
3 0

It looks like you want to compute the double integral

\displaystyle \iint_D (x+y) \,\mathrm dx\,\mathrm dy

over the region <em>D</em> with the unit circle <em>x</em> ² + <em>y</em> ² = 1 as its boundary.

Convert to polar coordinates, in which <em>D</em> is given by the set

<em>D</em> = {(<em>r</em>, <em>θ</em>) : 0 ≤ <em>r</em> ≤ 1 and 0 ≤ <em>θ</em> ≤ 2<em>π</em>}

and

<em>x</em> = <em>r</em> cos(<em>θ</em>)

<em>y</em> = <em>r</em> sin(<em>θ</em>)

d<em>x</em> d<em>y</em> = <em>r</em> d<em>r</em> d<em>θ</em>

Then the integral is

\displaystyle \iint_D (x+y)\,\mathrm dx\,\mathrm dy = \iint_D r^2(\cos(\theta)+\sin(\theta))\,\mathrm dr\,\mathrm d\theta \\\\ = \int_0^{2\pi} \int_0^1 r^2(\cos(\theta)+\sin(\theta))\,\mathrm dr\,\mathrm d\theta \\\\ = \underbrace{\left( \int_0^{2\pi}(\cos(\theta)+\sin(\theta))\,\mathrm d\theta \right)}_{\int = 0} \left( \int_0^1 r^2\,\mathrm dr \right) = \boxed{0}

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