The integral that is an equivalent iterated integral in three of the five possible other orders is
I= [tex]\int\limits^9_0[/tex] [tex]\int\limits^{9-z}_0[/tex][tex]\int\limits^3_{\sqrt{9-y}[/tex] f(x, y, z)dzdydx
An integral in mathematics assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that come from merging infinitesimal data. The process of determining integrals is known as integration.
Consider the integral,
I= [tex]\int\limits^3_0[/tex] [tex]\int\limits^9_{9-x^{2}}[/tex] [tex]\int\limits^{9-y}_0[/tex]f(x, y, z)dzdydx
This implies that the solid is bounded by
0≤z≤9-y
9-x²≤y≤9
0≤x≤3
Therefore, y+z=9, z=0
y=9-x², y=9
x=0, x=3
Now, the region is given by,
y=9-x², y=9, x=3, x=0
The shaded region is the required region of the figure region.
For the next integral,
√9-y≤x≤3
0≤y≤9
I= [tex]\int\limits^9_0[/tex] [tex]\int\limits^3_{\sqrt{9-y}[/tex] [tex]\int\limits^{9-y}_0[/tex]f(x, y, z)dzdydx
Here we have,
y+z =9, z=0
y=9-x², y=9
x=0, x=3
So, √9-y≤x≤3
The projection on the yz plane is,
y+z=0, z=0
y=9
0≤z≤9-y
0≤y≤9
I= [tex]\int\limits^9_0[/tex] [tex]\int\limits^{9-y}_0[/tex][tex]\int\limits^3_{\sqrt{9-y}[/tex] f(x, y, z)dzdydx
Again, 0≤y≤9-z
0≤z≤9
I= [tex]\int\limits^9_0[/tex] [tex]\int\limits^{9-z}_0[/tex][tex]\int\limits^3_{\sqrt{9-y}[/tex] f(x, y, z)dzdydx
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