Evaluate the integral $\int_{0}^{9} \int_{0}^{3} \int_{4 y}^{12} \frac{5 \cos \left(x^{2}\right)}{6 \sqrt{z}} d x d y d z$ by changing the order of integration in an appropriate way.
t: 0
\[
\int_{0}^{9} \int_{0}^{3} \int_{4 y}^{12} \frac{5 \cos \left(x^{2}\right)}{6 \sqrt{z}} d x d y d z=
\]
Final Answer: The value of the integral is \(\boxed{\frac{5\sin(144)}{8}}\).
Step 1 :The given triple integral is \(\int_{0}^{9} \int_{0}^{3} \int_{4 y}^{12} \frac{5 \cos \left(x^{2}\right)}{6 \sqrt{z}} d x d y d z\).
Step 2 :The limits of integration for x are functions of y, and the integrand is a function of x and z.
Step 3 :To change the order of integration, we need to consider the region of integration in the xyz-space. The region is defined by the inequalities 0 ≤ y ≤ 3, 0 ≤ z ≤ 9, and 4y ≤ x ≤ 12.
Step 4 :We can change the order of integration to dy dz dx by considering the projection of the region onto the yz-plane. The new limits of integration for y will be from 0 to x/4, for z from 0 to 9, and for x from 0 to 12.
Step 5 :The integrand is \(\frac{5\cos(x^{2})}{6\sqrt{z}}\).
Step 6 :The integral of the integrand is \(\frac{5\sin(144)}{8}\).
Step 7 :Final Answer: The value of the integral is \(\boxed{\frac{5\sin(144)}{8}}\).