Find the volume of the given solid.

Enclosed by the paraboloid
z = 6x^2 + 2y^2
and the planes
x = 0, y = 4, y = x, z = 0

81, guarantee

Well, let's see if we can solve this problem without getting too parabolic!

First, let's sketch what we have. We have a paraboloid and four planes. Now, look at the equation of the paraboloid, z = 6x^2 + 2y^2. This means that the paraboloid keeps getting bigger and bigger as x and y increase. So, we know that our solid is going to be bounded by the planes y = 4, y = x, and z = 0.

The plane x = 0 is actually just the yz-plane, so that's going to be our boundary on the left side. The plane y = 4 is a horizontal plane that's going to intersect the paraboloid at some point. And the plane y = x is actually a plane in 3D space that's going to intersect the paraboloid in the shape of a slanted circle.

Now, let's find the limits of integration for x, y, and z. For x, we can see that it goes from x = 0 to x = y, since y = x is one of our boundaries. For y, it goes from y = 0 to y = 4. And for z, it goes from z = 0 to z = 6x^2 + 2y^2.

Okay, now let's set up our integral:

∫∫∫ 1 dz dy dx

with the limits of integration as described above.

And there you have it! The volume of the given solid. Just don't go paraboling into it... wouldn't want you to get too quadratic!

To find the volume of the given solid enclosed by the paraboloid and the planes, we can use a triple integral.

First, let's find the limits of integration for each variable.

For x, the limits are given by the planes x = 0 and y = x, which means x goes from 0 to y.

For y, the limits are given by the planes y = 4 and y = x, which means y goes from x to 4.

For z, the limits are given by the paraboloid and the plane z = 0. Since the paraboloid is defined as z = 6x^2 + 2y^2, we can set z equal to this expression. Therefore, z goes from 0 to 6x^2 + 2y^2.

Now, we can set up the triple integral to find the volume:

Volume = ∫∫∫ dzdydx

∫(∫(∫ dz from 0 to 6x^2 + 2y^2)dy from x to 4)dx from 0 to y

Integrating the innermost integral with respect to z:

Volume = ∫(∫[6x^2 + 2y^2]dz from 0 to 6x^2 + 2y^2)dy from x to 4)dx from 0 to y

Simplifying the inner integrals:

Volume = ∫(6x^2 + 2y^2) (6x^2 + 2y^2 - 0) dy from x to 4)dx from 0 to y

Integrating the remaining integrals:

Volume = ∫(6x^2 + 2y^2) [(6x^2 + 2y^2) (4 - x)] dx from 0 to y

Simplifying and continuing the integration:

Volume = ∫(12x^4 + 4x^2y^2 + 12x^2y^2 + 4y^4 - 6x^6 - 2x^4y^2 - 6x^2y^4 - 2y^6) dx from 0 to y

Integrating:

Volume = [(12/5)x^5 + (4/3)x^3y^2 + (12/3)x^3y^2 + (4/5)yx^5 - (6/7)x^7 - (2/5)x^5y^2 - (6/5)x^3y^4 - (2/7)x^7y^2] from 0 to y

Simplifying:

Volume = (12/5)y^5 + (4/3)y^5 + (12/3)y^5 + (4/5)y^6 - (6/7)y^7 - (2/5)y^7 - (6/5)y^7 - (2/7)y^7

Simplifying further:

Volume = 12/5 y^5 + 4/3 y^5 + 4/3 y^6 - 6/7 y^7

Finally, we evaluate the integral at the limits of integration:

Volume = (12/5)(y^5/5) + (4/3)(y^5/5) + (4/3)(y^6/6) - (6/7)(y^7/7) evaluated from x=0 to y=4

Volume = (12/5)(4^5/5) + (4/3)(4^5/5) + (4/3)(4^6/6) - (6/7)(4^7/7) - [(12/5)(0^5/5) + (4/3)(0^5/5) + (4/3)(0^6/6) - (6/7)(0^7/7)]

Volume = (12/5)(1024/5) + (4/3)(1024/5) + (4/3)(4096/6) - (6/7)(16384/7)

Volume = (24576/25) + (8192/15) + (8192/9) - (221184/49)

Volume = 5904384/1225 + 98304/375 + 16384/225 - 4515072/2401

Volume ≈ 29.924

Therefore, the volume of the given solid is approximately 29.924 cubic units.

To find the volume of the solid enclosed by the paraboloid and the planes, we can use a triple integral.

Step 1: Determine the limits of integration for each variable.
To find the limits of integration, we need to determine the range of each variable within the given restrictions.

x = 0: This means x varies from 0 to the highest value it can take within the given region. Looking at the planes, we see that y = x. So x can take values from 0 to 4.

y = 4: This means y is fixed at 4 and does not vary.

y = x: This equation represents a line where y is equal to x. We already found that x varies from 0 to 4, so y also varies from 0 to 4.

z = 0: This equation represents the xy-plane, and z is fixed at 0.

Step 2: Set up the triple integral.
The volume V can be expressed as the triple integral of 1 with respect to x, y, and z over the given region.

V = ∫∫∫ 1 dV

Since the region is bounded by the planes x = 0, y = 4, y = x, and z = 0, the limits of integration are as follows:

∫(0 to 4)∫(0 to 4)∫(0 to 6x^2+2y^2) 1 dz dy dx

Step 3: Evaluate the triple integral.
Integrating the constant 1 over the given limits will yield the volume of the solid.

V = ∫∫∫ 1 dz dy dx
= ∫(0 to 4)∫(0 to 4)∫(0 to 6x^2+2y^2) 1 dz dy dx

Evaluating this integral will give you the volume of the solid enclosed by the paraboloid and the planes.