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

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

Enrhifo

$$\frac{\pi \left(\ln(\sqrt{5}+2)+2\sqrt{5}\right)\gamma }{2}$$

Camau Ateb

Gwerthuso’r integryn amhenodol yn gyntaf.

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

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

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

Symleiddio.

$$\frac{\left(2\sqrt{5}+\ln(2+\sqrt{5})\right)\gamma \theta }{4}$$

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.

$$\frac{1}{4}\left(2\times 5^{\frac{1}{2}}+\ln(2+5^{\frac{1}{2}})\right)\gamma \times 2\pi -\frac{1}{4}\left(2\times 5^{\frac{1}{2}}+\ln(2+5^{\frac{1}{2}})\right)\gamma \times 0$$

Symleiddio.

$$\frac{\left(2\sqrt{5}+\ln(2+\sqrt{5})\right)\gamma \pi }{2}$$

Gwahaniaethu w.r.t. γ

$$\frac{\pi \ln(\sqrt{5}+2)}{2}+\pi \sqrt{5}$$