Find the volume bounded above by parabolic z = 6 - x^2 - y^2 and below by z= 2x^2 + y^2

The paraboloids intersect along an ellipse

3x^2+2y^2 = 6

So, the volume is (using symmetry)

v = 4∫[0,√2] ∫[0,√((6-3x^2)/2)] (6-x^2-y^2)-(2x^2+y^2) dy dx = 3π√6

Evaluate double integral ln((x-y)/(x+y)) dy dx where the double integral region is the triangle with vertices (1,0),(4,3), (4,1). Hint: use a transformation with the Jacobian.

To find the volume bounded above by the parabolic surface z = 6 - x^2 - y^2 and below by the surface z = 2x^2 + y^2, we need to set up a double integral.

Step 1: Determine the limits of integration for x and y.

The parabolic surface z = 6 - x^2 - y^2 is a downward-opening paraboloid, while the surface z = 2x^2 + y^2 is an upward-opening elliptic paraboloid. To find the limits of integration, we need to find the region in the xy-plane where these two surfaces intersect.

Setting the two surfaces equal to each other:

6 - x^2 - y^2 = 2x^2 + y^2

Rearranging the equation:

3x^2 + 2y^2 = 6

Dividing through by 6:

x^2/2 + y^2/3 = 1

This equation represents an ellipse with semi-major axis along the y-axis of √3 and semi-minor axis along the x-axis of √2. Therefore, the limits of integration for x will be -√2 to √2, and for y, it will be -√3 to √3.

Step 2: Set up the double integral.

The volume can be calculated by integrating the height (z) over the region determined by the limits of integration for x and y. The integral will be set up as follows:

V = ∬R (6 - x^2 - y^2) - (2x^2 + y^2) dA

where R is the region in the xy-plane bounded by the ellipse x^2/2 + y^2/3 = 1.

Step 3: Evaluate the double integral.

Integrating the expression:

V = ∬R (4 - 3x^2 - 4y^2) dA

To evaluate the double integral, we can switch to polar coordinates. The equation of the ellipse in polar coordinates is:

r^2 = 2 cos^2(θ) + 3 sin^2(θ)

Since the ellipse has symmetry with respect to both the x-axis and the y-axis, we can consider integrating over one quadrant and then multiply the result by 4 to account for the other quadrants.

The limits of integration in polar coordinates will be:
-θ: 0 to π/2 (one quadrant)
r: 0 to √(2 cos^2(θ) + 3 sin^2(θ))

The volume can then be expressed as:

V = 4 ∫[0, π/2] ∫[0, √(2 cos^2(θ) + 3 sin^2(θ))] (4 - 3r^2) r dr dθ

Evaluating this integral will give you the final volume of the region bounded by the two surfaces.

To find the volume bounded above by the paraboloid z = 6 - x^2 - y^2 and below by the parabolic cylinder z = 2x^2 + y^2, we will use a triple integral.

First, we need to determine the region of integration. This can be done by finding the intersection curve of the two surfaces. Since both surfaces are quadratic, we can set them equal to each other:

6 - x^2 - y^2 = 2x^2 + y^2

Rearranging the equation, we get:

3x^2 + 2y^2 - 6 = 0

This equation represents an elliptical curve. To simplify it further, we can divide through by 6:

x^2/2 + y^2/3 = 1

This is the equation of an ellipse centered at the origin. Now we need to find the range of x and y values that lie within this ellipse.

To do this, we can parameterize the ellipse using polar coordinates:

x = √(2)cosθ
y = √(3)sinθ

Here, θ ranges from 0 to 2π to cover the entire ellipse.

Next, we can set up the triple integral to calculate the volume. Since the bounds for x and y have been determined, we only need to determine the range for z. The z range can be found by setting up the inequality:

2x^2 + y^2 ≤ 6 - x^2 - y^2

Rearranging, we get:

3x^2 + 2y^2 ≤ 6

Now we can set up the triple integral:

Volume = ∫∫∫ (dz dy dx)

with the following bounds:

x: -√(2)cosθ to √(2)cosθ
y: -√(3)sinθ to √(3)sinθ
z: 2x^2 + y^2 to 6 - x^2 - y^2

Integrating over these bounds will give us the volume bounded by the two surfaces.