Question

o valor da integral -> S 2 -> S 3 9x ²y ²dydx é:
                                 ->S-2 ->S 0

Escolha uma:
a) -234
b) 432
c) -432
d) 234
e) 0

obs : integral dupla o ->S representa.

Answer 1

Olá


Alternativa correta, letra B) 432


$\displaystyle \mathsf{ \int\limits^2_{-2} \, \int\limits^3_0 {9x^2y^2} \,dy dx }$


A primeira integral está em dy, por tanto quaisquer outras variáveis diferente de 'y' se tornam constantes.


$\displaystyle \mathsf{ \int\limits^2_{-2} \, \int\limits^3_0 {9x^2y^2} \,dy dx }\\\\\\\displaystyle \mathsf{ \int\limits^2_{-2} \, \left[9x^2\int\limits^3_0 {y^2} \,dy\right] dx }\\\\\\\displaystyle \mathsf{ \int\limits^2_{-2} \, \left[ {9x^2 \cdot \left(\frac{y^{2+1}}{2+1}\right)\bigg|^3_0 } \,\right] dx }\\\\\\\displaystyle \mathsf{ \int\limits^2_{-2} \, \left[ {9x^2 \cdot \left(\frac{3^{3}}{3}~-~ \frac{0^3}{3} \right)} \,\right] dx }$

$\displaystyle \\\\\\\displaystyle \mathsf{ \int\limits^2_{-2} \, \left[ {9x^2 \cdot \frac{27}{3} } \,\right] dx }\\\\\\ \mathsf{ \int\limits^2_{-2} \, 81x^2\, dx }}\\\\\\\\\mathsf{81\cdot \left( \frac{x^{2+1}}{2+1}\right)\bigg|^2_{-2}}\\\\\\\\\mathsf{81\cdot \left( \frac{2^3}{3}~-~ \frac{(-2)^3}{3} \right)}\\\\\\\\\mathsf{81\cdot \frac{16}{3} }\\\\\\\\\boxed{\mathsf{432}}\qquad\qquad\Longrightarrow\qquad\text{Letra B)}$