Find the area of the surface. The part of the paraboloid z = 1 - x2 - y2 that lies above the plane z = -6 ​

Answer:

π/6 [(29)^³/₂ − 1]

81.247

Step-by-step explanation:

Surface area is:

S = ∫∫ √(fₓ² + fᵧ² + 1) dA

The partial derivatives are:

fₓ = -2x

fᵧ = -2y

Substituting:

S = ∫∫ √(4x² + 4y² + 1) dA

To make this easier, we can convert to polar coordinates.

S = ∫∫ √(4r² + 1) dA

S = ∫∫ √(4r² + 1) r dr dθ

S = ⅛ ∫∫ 8r √(4r² + 1) dr dθ

The limits for θ are 0 to 2π.  The minimum of r is 0.  The maximum of r is:

-6 = 1 − x² − y²

-6 = 1 − r²

r² = 7

r = √7

Integrating the first integral:

S = ⅛ ∫ ⅔ (4r² + 1)^³/₂ |₀ᴿ dθ

S = ⅛ ∫ [⅔ (4(7) + 1)^³/₂ − ⅔ (4(0) + 1)^³/₂] dθ

S = ⅛ ∫ [⅔ (29)^³/₂ − ⅔] dθ

S = ¹/₁₂ [(29)^³/₂ − 1] ∫ dθ

Integrating the second integral:

S = ¹/₁₂ [(29)^³/₂ − 1] θ |₀²ᵖⁱ

S = ¹/₁₂ [(29)^³/₂ − 1] (2π − 0)

S = π/6 [(29)^³/₂ − 1]

S ≈ 81.247

The area of a shape is the amount of space it covers.

The surface area is approximately 81.3 unit squares

The given parameters are:

[tex]\mathbf{z = 1 - x^2 - y^2}[/tex]

[tex]\mathbf{z = -6}[/tex]

Calculate the partial derivatives of [tex]\mathbf{z = 1 - x^2 - y^2}[/tex]

[tex]\mathbf{f_x = -2x}[/tex]

[tex]\mathbf{f_y = -2y}[/tex]

So, the surface area is:

[tex]\mathbf{A = \int\limits^a_b\int\limits^a_b {\sqrt{(f_x^2 + f_y^2 +1 )}} \, dA }[/tex]

So, we have:

[tex]\mathbf{A = \int\limits^a_b\int\limits^a_b {\sqrt{((-2x)^2 + (-2y)^2 +1 )}} \, dA }[/tex]

[tex]\mathbf{A = \int\limits^a_b\int\limits^a_b {\sqrt{(4x^2 + 4y^2 +1 )}} \, dA }[/tex]

Factor out 4

[tex]\mathbf{A = \int\limits^a_b\int\limits^a_b {\sqrt{(4(x^2 + y^2) +1 )}} \, dA }[/tex]

Substitute

[tex]\mathbf{r^2 = x^2 + y^2}\\\mathbf{dA = rdr d\theta}[/tex]

So, we have:

[tex]\mathbf{A = \int\limits^a_b\int\limits^a_b {\sqrt{(4r^2 +1 )}} \, r\ dr\ d\theta }[/tex]

Multiply by 1

[tex]\mathbf{A = \int\limits^a_b\int\limits^a_b {\sqrt{(4r^2 +1 )}} \, r \times 1\ dr\ d\theta }[/tex]

Express 1 as 8/8

[tex]\mathbf{A = \int\limits^a_b\int\limits^a_b {\sqrt{(4r^2 +1 )}} \, r \times \frac{8}{8}\ dr\ d\theta }[/tex]

Rewrite as:

[tex]\mathbf{A =\frac{1}{8} \int\limits^a_b\int\limits^a_b {8r \sqrt{(4r^2 +1 )}}, \ dr\ d\theta }[/tex]

From the question,

[tex]\mathbf{z = -6}[/tex]

So, we have:

[tex]\mathbf{1 - r^2 = -6}[/tex]

Add 6 to both sides

[tex]\mathbf{7 - r^2 = 0}[/tex]

So, we have:

[tex]\mathbf{r^2 = 7}[/tex]

Integrate [tex]\mathbf{A =\frac{1}{8} \int\limits^a_b\int\limits^a_b {8r \sqrt{(4r^2 +1 )}}, \ dr\ d\theta }[/tex]

[tex]\mathbf{A =\frac{1}{8} \int\limits^a_b {\frac{8}{12} (4r^2 +1 )^{\frac 32}}, |\limits^R_0 d\theta }[/tex]

[tex]\mathbf{A =\frac{1}{8} \times \frac{8}{12} \int\limits^a_b { (4r^2 +1 )^{\frac 32}}, |\limits^R_0 d\theta }[/tex]

[tex]\mathbf{A =\frac{1}{12} \int\limits^a_b { (4r^2 +1 )^{\frac 32}}, |\limits^R_0 d\theta }[/tex]

[tex]\mathbf{A =\frac{1}{12} \int\limits^a_b { (4r^2 +1 )^{\frac 32} -(4 \times 0^2 +1 )^{\frac 32} }, d\theta }[/tex]

Substitute [tex]\mathbf{r^2 = 7}[/tex]

[tex]\mathbf{A =\frac{1}{12} \int\limits^a_b { (4\times 7 +1 )^{\frac 32} - 1^{\frac 32}}, d\theta }[/tex]

[tex]\mathbf{A =\frac{1}{12} \int\limits^a_b {29^{\frac 32} - 1}, d\theta }[/tex]

Rewrite as:

[tex]\mathbf{A =\frac{1}{12} (29^{\frac 32} - 1), \int\limits^a_b d\theta }[/tex]

Integrate

[tex]\mathbf{A =\frac{1}{12} (29^{\frac 32} - 1), \times \theta |\limits^{2\pi}_0}[/tex]

So, we have:

[tex]\mathbf{A =\frac{1}{12} (29^{\frac 32} - 1), \times (2\pi - 0) }[/tex]

[tex]\mathbf{A =\frac{1}{12} (29^{\frac 32} - 1), \times 2\pi}[/tex]

Rewrite as:

[tex]\mathbf{A =\frac{2\pi}{12} (29^{\frac 32} - 1)}[/tex]

[tex]\mathbf{A =\frac{\pi}{6} (29^{\frac 32} - 1)}[/tex]

[tex]\mathbf{A =81.3}[/tex]

Hence, the surface area is approximately 81.3 unit squares

Read more about surface areas at:

https://brainly.com/question/3621496

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