Final answer:
The volume of the region under the surface z=3x^4 and above the given triangle can be found by calculating the cross-sectional area of the triangle and multiplying it by the height of the region.
Step-by-step explanation:
To find the volume of the region under the surface z=3x^4 and above the triangle in the xy-plane with corners (0,0,0), (6,0,0) and (0,5,0), we need to calculate the cross-sectional area of the triangle and multiply it by the height of the region.
The cross-sectional area of a triangle can be found using the formula A = 1/2 * base * height. In this case, the base of the triangle is 6 units and the height is 5 units, so the area of the triangle is A = 1/2 * 6 * 5 = 15 square units.
The height of the region is given by the z-coordinate of the surface z=3x^4. Since the surface is flat and parallel to the xy-plane, the height is constant and equal to 3 times the maximum value of x^4 within the triangle.
To find the maximum value of x^4 within the triangle, we can evaluate x^4 at each of the three vertices of the triangle. Evaluating x^4 at (0,0,0) gives 0, evaluating it at (6,0,0) gives 6^4 = 1296, and evaluating it at (0,5,0) gives 0. Therefore, the maximum value of x^4 within the triangle is 1296.
The height of the region is then 3 times 1296, which is 3888. Multiplying the area of the triangle by the height gives us the volume of the region: V = 15 * 3888 = 58320 cubic units.