Here is the region of integration of the integral integral from negative 6 to 6 integral from x squared to 36 integral from 0 to 36 minus y dz dy dx. rewrite the integral as an equivalent integral in the following orders. a. dy dz dx by. dy dx dz c. dx dy dz d. dx dz dy e. dz dx dy

a) ∫_{-6}^{6} ∫_{0}^{36} ∫_{x²}^{36} (-y) dy dz dx

b)  ∫_{0}^{36} ∫_{-6}^{6} ∫_{x²}^{36} (-y) dy dx dz

c)  ∫_{0}^{36} ∫_{x²}^{36} ∫_{-6}^{6} (-y) dx dy dz

e) ∫_{x²}^{36} ∫_{-6}^{6} ∫_{0}^{36} (-y) dz dx dy

Step-by-step explanation:

We write the equivalent integrals for given integral,

we get:

a) ∫_{-6}^{6} ∫_{0}^{36} ∫_{x²}^{36} (-y) dy dz dx

b)  ∫_{0}^{36} ∫_{-6}^{6} ∫_{x²}^{36} (-y) dy dx dz

c)  ∫_{0}^{36} ∫_{x²}^{36} ∫_{-6}^{6} (-y) dx dy dz

e) ∫_{x²}^{36} ∫_{-6}^{6} ∫_{0}^{36} (-y) dz dx dy

We changed places of integration, and changed boundaries for certain integrals.

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation: