Réitigh ∫ (from 0 to 2 pi) of int_0^1rsqrt{4r^2+1}drdθ

$$\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 1 } r \sqrt { 4 r ^ { 2 } + 1 } d r d \theta$$

Luacháil

$$\frac{5\pi \sqrt{5}-\pi }{6}\approx 5.3304135$$

Céimeanna Réitigh

Déan luacháil ar an suimeálaí éiginnte ar dtús.

$$\int \int _{0}^{1}r\sqrt{4r^{2}+1}\mathrm{d}r\mathrm{d}\theta$$

Aimsigh suimeálaithe do \int _{0}^{1}r\sqrt{4r^{2}+1}\mathrm{d}r ag baint úsáid as an tábla do suimeálaithe coitianta riail\int a\mathrm{d}\theta =a\theta .

$$\int _{0}^{1}r\sqrt{4r^{2}+1}\mathrm{d}r\theta$$

Simpligh.

$$\frac{5\sqrt{5}-1}{12}\theta$$

Is ionann suimeálaí cinnte agus frithdhíorthach an nath luacháilte ag teorainn uachtair na suimeála lúide an frithdhíorthach luacháilte ag teorainn íochtair na suimeála.

$$\left(\frac{5}{12}\times 5^{\frac{1}{2}}-\frac{1}{12}\right)\times 2\pi -\left(\frac{5}{12}\times 5^{\frac{1}{2}}-\frac{1}{12}\right)\times 0$$

Simpligh.

$$\frac{5\sqrt{5}\pi -\pi }{6}$$