Question

Find the surface area of the surface S, where S is the portion of the surface $6\sqrt{15x + 6y^2 - 12\ln y - 6z} = 0$ that lies above the rectangle $0 \le x \le 1$ and $1 \le y \le 9$ in the xy-plane A. $\frac{1}{3}[40 + \ln 9]$ B. $4[40 + \ln 9]$ C. $2[40 + \ln 9]$ D. $12[40 + \ln 9]$

          Find the surface area of the surface S, where S is the portion of the surface $6\sqrt{15x + 6y^2 - 12\ln y - 6z} = 0$ that lies above the rectangle $0 \le x \le 1$ and $1 \le y \le 9$ in the xy-plane

A. $\frac{1}{3}[40 + \ln 9]$
B. $4[40 + \ln 9]$
C. $2[40 + \ln 9]$
D. $12[40 + \ln 9]$

Find the surface area of the surface S, where S is the portion of the surface 6√(15x + 6y^2 - 12ln y - 6z) = 0 that lies above the rectangle 0 ≤ x ≤ 1 and 1 ≤ y ≤ 9 in the xy-plane

A. (1)/(3)[40 + ln 9]
B. 4[40 + ln 9]
C. 2[40 + ln 9]
D. 12[40 + ln 9]

Added by Juan Antonio S.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Arun Bana

03/15/2022

Step 1

The given rectangle in the xy-plane is 0 ≤ x ≤ 1 and 0 ≤ y ≤ 9. Show more…

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Submittest point(s)possible Find the surface area of the surface S, where S is the portion of the surface 6T5x+6y12 In y-6z=0 that lies above the rectangle0x1and1y9 in thexy-plane OA. [40 + In 9] O B. 4[40 + In 9] O C. 2[40 + In 9] O D. 12[40 + In 9]