Question

Find the volume of the solid which lies below the surface f(x,y)=12x^3 and above the planar region in the x−y plane deseribed by 3x2+1⩽y⩽4, 0⩽x⩽1.

Answer 1

To find the volume of the solid below the surface f(x,y)=12x^3 and above the planar region in the x-y plane described by 3x^2+1<=y<=4, 0<=x<=1, set up a double integral over the region.

Step-by-step explanation:

To find the volume of the solid below the surface

f(x,y) = 12x^3

and above the planar region in the x-y plane described by

3x^2 + 1 ≤ y ≤ 4

and

0 ≤ x ≤ 1

we can set up a double integral over the region. The volume can be found using the formula:

V = ∫∫R f(x,y) dA

where R represents the region of integration, f(x,y) is the function defining the solid, and dA is the differential area element.

In this case, the region of integration is a rectangle in the x-y plane bounded by 0 ≤ x ≤ 1 and 3x^2 + 1 ≤ y ≤ 4. The differential area element dA can be expressed as dA = dx dy.

Therefore, the volume of the solid is:

V = ∫0 1 ∫3x^2 + 1 4 12x^3 dx dy

By evaluating this double integral, you can find the volume of the solid.