Question

Determine a centroide da região solida delimitada inferiormente pela paraboloide Z= 3(x²+y²) = 3r² e superiormente pelo plano Z=5. Represente por V o volume dessa região solida.

Answer 1

$3r^2=5\\\\ r= \sqrt{\frac{3}{3} }$

paraboloid

3r² ≤ z ≤ 5
0 ≤ r ≤ √(5/3)
0 ≤Ф≤ 2pi




$centroid = \frac{\int_{V} z\;dV}{V} = \frac{1}{V} \left [ \int\limits^{2\pi}_0 \int\limits_0^ {\sqrt{\frac{5}{3}}} \left[\int_{3r^2}^5 z\;dz \right ] r\;dr\;d\theta \right ]\\\\entroid = \frac{1}{V} \left [ \int\limits^{2\pi}_0 \int\limits_0^ {\sqrt{\frac{5}{3}}} \left[ \frac{25}{2}- \frac{9r^4}{2} \right ] r\;dr\;d\theta \right ]$

$centroid = \frac{1}{2V} \left [ \int\limits^{2\pi}_0 \int\limits_0^ {\sqrt{\frac{5}{3}}} (25r-9r^5)\;dr\;d\theta \right ]\\\\\\ centroid = \frac{1}{2V} \left [ \int\limits^{2\pi}_0d\theta * \int\limits_0^ {\sqrt{\frac{5}{3}}} (25r-9r^5)\;dr \right ]\\\\ centroid = \frac{1}{2V}\left[ 2\pi * \left[ \frac{25*( \sqrt{ \frac{5}{3} })^2 }{2}- \frac{9*( \sqrt{ \frac{5}{3} } )^6}{6} \right] \right] \\\\ \boxed{\boxed{centroid = \frac{1}{2V}\left[ 2\pi * \frac{125}{9} \right] = \frac{125\pi}{9V} }}$

Answer 2

Resposta:

125pi/9V

Explicação passo-a-passo: