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

[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

 

[HallsofIvy]

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.)

 

[tacman]

To find the volume of a solid below the surface z = f(x,y) and above the region in the xy plane, we can use the formula V = double integral_R f(x,y)dA, where R is the region in the xy plane and f(x,y) is the function defining the surface. In this case, our region R is defined by the boundaries x = 0, x = PI, y = 0, and y = PI.

To solve this integral, we can first integrate with respect to y, treating x as a constant. This gives us:

V = integral PI to 0 (integral PI to 0 (3 + cos(x) + cos(y))dy)dx

Integrating with respect to y, we get:

V = integral PI to 0 (3y + sin(y) + cos(x)y) from y = 0 to y = PI dx

Evaluating this integral, we get:

V = integral PI to 0 (3PI + sin(PI) + cos(x)PI - 0 - sin(0) - cos(x)0) dx

Simplifying this, we get:

V = integral PI to 0 (3PI + cos(x)PI) dx

Integrating with respect to x, we get:

V = (3PIx + sin(x)PI) from x = 0 to x = PI

Plugging in the limits of integration, we get:

V = 3PI^2 + sin(PI)PI - (0 + sin(0)0)

Simplifying this, we get the final answer:

V = 3PI^2 + 0 - 0 = 3PI^2

Therefore, the volume of the solid is 3PI^2 units cubed. I hope this helps clarify the process for solving this type of problem. Good luck with your studies!