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

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

Baholash

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

Yechim qadamlari

Evaluate the indefinite integral first.

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

Find the integral of \int _{0}^{1}\gamma \sqrt{4r^{2}+1}\mathrm{d}r using the table of common integrals rule \int a\mathrm{d}\theta =a\theta .

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

Qisqartirish.

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

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

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

Qisqartirish.

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

γ ga nisbatan hosilani topish

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