Respuesta :
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
Answer:
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
The surface area of that part of the plane 10x+7y+z=4 that lies inside the elliptic cylinder [tex]\dfrac{x^2}{25}+\dfrac{y^2}{9}=1[/tex] is [tex]15\pi\sqrt{150}[/tex] and this can be determined by using the given data.
Given :
- 10x + 7y + z = 4 ---- (1)
- [tex]\dfrac{x^2}{25}+\dfrac{y^2}{9}=1[/tex] --- (2)
Equation (1) can also be written as:
z = 4 - 10x - 7y ---- (3)
The surface area is given by the equation:
[tex]\rm A(s) = \int \int \sqrt{(\dfrac{\delta f}{\delta x})^2+(\dfrac{\delta f}{\delta y})^2+1}\;dA[/tex] --- (4)
[tex]\dfrac{\delta f}{\delta x} = -10[/tex]
[tex]\dfrac{\delta f}{\delta y} = -7[/tex]
Now, substitute the known values in the equation (4).
[tex]\rm A(s) = \int \int \sqrt{(10)^2+(7)^2+1}\;dA[/tex]
[tex]\rm A(s) = \sqrt{150} \int \int\;dA[/tex]
Now the area enclosed by an ellipse is given by:
[tex]\dfrac{x^2}{25}+\dfrac{y^2}{9}=1[/tex]
[tex]\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1[/tex]
By comparing the above equation:
a = 5
b = 3
The area is given by:
[tex]\rm A(s)=\sqrt{150}\times \pi(5\times 3)[/tex]
[tex]\rm A(s) = 15\pi \sqrt{150}[/tex]
For more information, refer to the link given below:
https://brainly.com/question/11952845
