Find the surface area of (x - 7)² + (y + 1)² + (z - 4)² = 4 that lies below z = 6. A) 8 $\pi$ B) 32 $\pi$ C) 16 $\pi$ D) 64 $\pi$ E) 128 $\pi$

Question

Quizplus · Accepted Answer

First, we need to find the intersection between the given surface and the plane z=6. (x - 7)² + (y + 1)² + (z - 4)² = 4 (z - 4)² = 4 - (x - 7)² - (y + 1)² z = 4/(²) - (x - 7)/(²) - (y + 1)/(²) + 4 Now we substitute this expression for z into our equation of the surface and get: (x - 7)² + (y + 1)² + (4/(²) - (x - 7)/(²) - (y + 1)/(²) + 4)² = 4 Simplifying, we get: 5(x - 7)/(²) + 5(y + 1)/(²) = -56 This is the equation of a plane that passes through the point (7,-1,6) and is parallel to the given surface. Now we need to find the intersection between this plane and the plane z=6. If we solve the system of equations: 5(x - 7)/(²) + 5(y + 1)/(²) = -56 z = 6 We get: x = 7 - (18²)/5 y = -1 + (34²)/5 Now we can find the gradient vector of the given surface at the point (x,y,z) and calculate its magnitude as follows: grad((x - 7)² + (y + 1)² + (z - 4)²) = <², ², ²> |grad((x - 7)² + (y + 1)² + (z - 4)²)| = ²√3 Therefore, the area of the portion of the surface that lies below z=6 is: Area = ∫∫|grad((x - 7)² + (y + 1)² + (z - 4)²)| dA over the region R where R is the projection of the portion of the surface that lies below z=6 onto the xy-plane. R is just a circle with center (7,-1) and radius |z-6| = 2. Therefore, Area = ∫∫²√3 dA over the circle of radius 2 centered at (7,-1) Using polar coordinates, we can compute this integral and get: Area = 16S1F1F1S1F1F1F10 = 16F1F1F1S1F1F1F10, which is choice C.

Find the Surface Area of (X - 7)2 + (Y π\piπ

Find the Surface Area of (X - 7)² + (Y $\pi$