How Do You Calculate the Volume of a Solid Defined by z = 3 + cos(x) + cos(y)?

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SUMMARY

The volume of the solid defined by the surface z = 3 + cos(x) + cos(y) above the region in the xy-plane bounded by x = 0, x = π, y = 0, and y = π is calculated using a double integral. The correct setup for the volume V is V = ∫ from 0 to π (∫ from 0 to π (3 + cos(x) + cos(y)) dy) dx. The final computed volume is 3π², confirming that the function remains above the xy-plane throughout the specified region.

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Dx
3) Find the volume of the solid that lies below the surface z = f(x,y) and above region in xy plane: z = 3+cos(x) + cos(y); x = 0; x = PI; y = 0; y=PI.

V=double integral_R f(x,y)dA; f(x,y)= 3 + cos(x) + cos(y); 0<= x <= PI and 0 <= PI so V = integral PI ro 0 (integral PI to 0 (3 + cos(x) + cos(y))dy)dx = ?

I am using a example in my book but am stuck here or confused if I am going in the right direction. please help?
Thanks!
Dx :wink:
 
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The only comment I would make is that it would make more sense to integrate from 0 to pi than from pi to 0!

Of course, since cos(x) and cos(y) are never less than -1,
3+ cos(x)+ cos(y) is never 0 so the function surface is always above the x,y plane.

Now, go ahead and do the integral. (I get 3pi2.)
 

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