Find the surface area of that part of the plane 10x+7y+z=4 that lies inside the elliptic cylinder x225+y29=1

The surface area of that part of the plane 10x+7y+z=4 that lies inside the elliptic cylinder is and this can be determined by using the given data.

Given :

Equation (1) can also be written as:

z = 4 - 10x - 7y      ---- (3)

The surface area is given by the equation:

    --- (4)

Now, substitute the known values in the equation (4).

Now the area enclosed by an ellipse is given by:

By comparing the above equation:

a = 5

b = 3

The area is given by:

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Correct question is;

Find the surface area of that part of the plane 10x + 7y + z = 4 that lies inside the elliptic cylinder x²/25 + y²/9 = 1

A(S) = 15π√150

Step-by-step explanation:

We are given;

10x + 7y + z = 4

Making z the subject, we have;

z = 4 - 10x - 7y

Now, area of the surface as part of z = f(x, y) is;

A(S) = ∫∫√[(∂f/∂x)² + (∂f/∂y)² + 1]dA

From z = 4 - 10x - 7y,

∂f/∂x = -10

∂f/∂y = -7

Thus;

A(S) = ∫∫√[(-10)² + (-7)² + 1]dA

A(S) = √150 ∫∫dA

Where ∫∫dA is the elliptical cylinder

From the general form of an area enclosed by an ellipse with the formula;

x²/a² + y²/b² = 1 and comparing with

x²/25 + y²/9 = 1, we have;

a = 5 and b = 3

So, area of elliptical cylinder = πab

Thus;

A(S) = √150 × π(5 × 3)

A(S) = 15π√150

only 10/60 isn't equivalent

hope this : )

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step-by-step explanation: