Question-215754

Question Number 215754 by universe last updated on 17/Jan/25

Answered by mr W last updated on 17/Jan/25

z=1+sin θ with −(π/2)≤θ≤(π/2) x=r cos φ, y=r sin φ with 0≤φ≤2π I=2π∫_(−(π/2)) ^(π/2) cos θ∫_0 ^(cos θ) [r^2 +(1+sin θ)^2 ]^(5/2) rdrdθ =((2π)/7)∫_(−(π/2)) ^(π/2) [2^(7/2) (1+sin θ)^(7/2) −(1+sin θ)^7 ]cos θdθ =((2π)/7)[2^(7/2) ×(2/9)(1+sin θ)^(9/2) −(1/8)(1+sin θ)^8 ]_(−(π/2)) ^(π/2) =((2π)/7)(2^(7/2) ×(2/9)×2^(9/2) −(2^8 /8)) =((64π)/9) ✓

$${z}=\mathrm{1}+\mathrm{sin}\:\theta\:{with}\:−\frac{\pi}{\mathrm{2}}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}} \\ $$$${x}={r}\:\mathrm{cos}\:\phi,\:{y}={r}\:\mathrm{sin}\:\phi\:{with}\:\mathrm{0}\leqslant\phi\leqslant\mathrm{2}\pi \\ $$$${I}=\mathrm{2}\pi\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\:\theta\int_{\mathrm{0}} ^{\mathrm{cos}\:\theta} \left[{r}^{\mathrm{2}} +\left(\mathrm{1}+\mathrm{sin}\:\theta\right)^{\mathrm{2}} \right]^{\frac{\mathrm{5}}{\mathrm{2}}} {rdrd}\theta \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{7}}\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \left[\mathrm{2}^{\frac{\mathrm{7}}{\mathrm{2}}} \left(\mathrm{1}+\mathrm{sin}\:\theta\right)^{\frac{\mathrm{7}}{\mathrm{2}}} −\left(\mathrm{1}+\mathrm{sin}\:\theta\right)^{\mathrm{7}} \right]\mathrm{cos}\:\theta{d}\theta \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{7}}\left[\mathrm{2}^{\frac{\mathrm{7}}{\mathrm{2}}} ×\frac{\mathrm{2}}{\mathrm{9}}\left(\mathrm{1}+\mathrm{sin}\:\theta\right)^{\frac{\mathrm{9}}{\mathrm{2}}} −\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+\mathrm{sin}\:\theta\right)^{\mathrm{8}} \right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{7}}\left(\mathrm{2}^{\frac{\mathrm{7}}{\mathrm{2}}} ×\frac{\mathrm{2}}{\mathrm{9}}×\mathrm{2}^{\frac{\mathrm{9}}{\mathrm{2}}} −\frac{\mathrm{2}^{\mathrm{8}} }{\mathrm{8}}\right) \\ $$$$=\frac{\mathrm{64}\pi}{\mathrm{9}}\:\checkmark \\ $$

Commented by mr W last updated on 17/Jan/25