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
