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

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

Aromātai

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

Ngā Upane Otinga

Aromātaitia te tau tōpū tautuhi-kore i te tuatahi.

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

Kimihia te tau tōpū o \int _{0}^{1}r\sqrt{4r^{2}+1}\mathrm{d}r mā te whakamahi i te ture mō te ripanga o ngā tau tōpū pātahi \int a\mathrm{d}\theta =a\theta .

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

Whakarūnātia.

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

Ko te tau tōpū tautuhi ko te pārōnaki kōaro o te kīanga i aromātaitia i te tepe tōrunga o te pāwhaitua, tangohia te pārōnaki kōaro i aromātaitia i te tepe tōraro o te pāwhaitua.

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

Whakarūnātia.

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