Find the volume of the given solid.
Enclosed by the paraboloid
z = 6x2 + 2y2
and the planes
x = 0, y = 4, y = x, z = 0
To find the volume of the solid enclosed by the paraboloid and the given planes, we need to set up a triple integral.
Let's start by graphing the region in the xy-plane that is enclosed by the planes x = 0, y = 4, and y = x.
The region is a triangle with vertices at (0,0), (4,4), and (4,0).
Next, we need to find the limits of integration for the triple integral.
Since the region is bounded by x = 0 and y = x, the x values range from 0 to 4.
Since the region is also bounded by y = 4, the y values range from 0 to 4.
Finally, we need to find the limits of z. The paraboloid equation, z = 6x^2 + 2y^2, tells us that z is always greater than or equal to 0. So the limits of z are from 0 to the height of the paraboloid.
To find the height, we need to find the maximum value of z within the region. Let's substitute y = x into the equation of the paraboloid:
z = 6x^2 + 2(x^2)
z = 8x^2
To find the maximum value of z, we take the derivative of z with respect to x and set it equal to 0.
dz/dx = 16x
16x = 0
x = 0
So the maximum value of z occurs when x = 0. Substituting this back into the equation of the paraboloid, we get:
z = 8(0)^2
z = 0
Therefore, the height of the paraboloid within the region is 0.
Now we can set up the triple integral:
∫∫∫ 1 dz dy dx
With the limits of integration:
0 ≤ x ≤ 4
0 ≤ y ≤ x
0 ≤ z ≤ 0
Notice that since the height of the paraboloid within the region is 0, the triple integral is reduced to a single point and the volume of the solid is equal to 0.
Therefore, the volume of the given solid is 0.