Question

Find the volume of the solid under the plane in the first octant z=3xy+y and above the region determined by y=x⁷ and y=x.

Answer 1

The volume of the solid under the plane z = 3xy + y and above the region determined by y = x^7 and y = x is approximately 14.15 cubic units.

Step-by-step explanation:

To find the volume of the solid, we need to integrate the function representing the height of the solid over the region determined by the inequalities. The height of the solid is given by z = 3xy + y, which is a function of x and y. The region determined by the inequalities is in the first octant, where both x and y are positive.

First, we find the boundary curves of the region. The curve y = x^7 intersects the curve y = x at x = 1, where both curves coincide. Therefore, we can integrate over this interval. The limits of integration for x are 0 to 1.

Next, we need to find the limits of integration for y over each curve. For y = x^7, we have 0 <= x <= 1, so 0 <= y <= x^7. For y = x, we have 0 <= x <= 1, so 0 <= y <= x.

Now, we can integrate over this region using double integration:

V = ∬∬ (3xy + y) dx dy

Substituting the limits of integration:

V = ∬∬ (3xy^2 + y^2) dx dy

Evaluating the integral:

`V = [(3/14)x^2y^3 + (1/6)y^3]_0^1 [y^2]_0^x^7`

Simplifying:

`V = (3/14)(x^9) + (1/6)(x^10) - (x^7)/6 |_0^1`

Evaluating:

V ≈ 14.15 cubic units.