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

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

Enrhifo

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

Camau Ateb

Gwerthuso’r integryn amhenodol yn gyntaf.

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

Canfod integryn \int _{0}^{1}r\sqrt{4r^{2}+1}\mathrm{d}r gan ddefnyddio'r rheol tabl o integrynnau cyffredin \int a\mathrm{d}\theta =a\theta .

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

Symleiddio.

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

Yr integryn pendant yw integryn amhendant y mynegiant wedi’i werthuso ar lefel uchaf yr integreiddiad llai’r integryn amhendant ar lefel isaf yr integreiddiad.

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

Symleiddio.

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