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Monica [59]
3 years ago
7

Tính tích phân sau bằng cách dùng tọa độ cực I=∫∫

y%5E%7B2%7D%20%7D%20%7D" id="TexFormula1" title="\frac{1}{\sqrt{x^{2} +y^{2} } }" alt="\frac{1}{\sqrt{x^{2} +y^{2} } }" align="absmiddle" class="latex-formula">dxdy R là miền nằm trọg góc phần tư thứ nhất thỏa mãn 4\leq x^{2} +y^{2} \leq 9
Mathematics
1 answer:
xenn [34]3 years ago
5 0

It sounds like <em>R</em> is the region (in polar coordinates)

<em>R</em> = {(<em>r</em>, <em>θ</em>) : 2 ≤ <em>r</em> ≤ 3 and 0 ≤ <em>θ</em> ≤ <em>π</em>/2}

Then the integral is

\displaystyle \iint_R\frac{\mathrm dx\,\mathrm dy}{\sqrt{x^2+y^2}} = \int_0^{\pi/2}\int_2^3 \frac{r\,\mathrm dr\,\mathrm d\theta}{\sqrt{r^2}} \\\\ = \int_0^{\pi/2}\int_2^3 \mathrm dr\,\mathrm d\theta \\\\ = \frac\pi2\int_2^3 \mathrm dr \\\\ = \frac\pi2r\bigg|_2^3 = \frac\pi2 (3-2) = \boxed{\frac\pi2}

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