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Integrate f(x,y,z) = x over the region w in the first octant above z=y² and below z=48-3x²-2y²?

User Gregh
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Final answer:

To integrate the function f(x,y,z) = x over the given region, you can use triple integration. Find the limits of integration for x, y, and z based on the given boundaries. Then set up and solve the triple integral.

Step-by-step explanation:

To integrate the function f(x,y,z) = x over the given region, we can use the concept of triple integration. First, let's find the limits of integration for x, y, and z. From the given information, we have z = y² as the lower boundary of the region and z = 48-3x²-2y² as the upper boundary. We can rewrite the upper boundary as z = 48 - (3x² + 2y²). So, the limits of integration for z are y² to 48 - (3x² + 2y²).

Next, we need to determine the limits of integration for y. Since the region is in the first octant, y starts from 0 and goes up to the curve y = sqrt(x). Therefore, the limits of integration for y are 0 to sqrt(x).

Finally, for x, we need to find the bounds of integration. The region is in the first octant, so x starts from 0 and goes up to the curve x = sqrt(48 - 2y²). Therefore, the limits of integration for x are 0 to sqrt(48 - 2y²).

Now, we can write the triple integral as follows:

∫∫∫ f(x,y,z) dV = ∫∫∫ x dz dy dx

= ∫∫ [0 to sqrt(48 - 2y²)] [y² to 48 - (3x² + 2y²)] x dz dy dx

= ∫ [0 to sqrt(48 - 2y²)] ∫ [y² to 48 - (3x² + 2y²)] x (48 - (3x² + 2y²) - y²) dz dy dx

Now, you can simplify and solve this triple integral to find the value of the integral.

User Jaspinder Kaur
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