Respuesta :
Answer:
A = 339.531 units^2
Step-by-step explanation:
Given:-
- The surface of consideration is bounded by the following:
Cylinder : [tex]Y = \frac{2}{3}*Z^\frac{3}{2}[/tex]
Planes : x = 3 , y = 18
Find:-
Find the area of the surface in the first octant
Solution:-
- It is better to sketch the given curves. Taking (y-z) plane on the sketch paper with x - axis out of the page.
- We see that a surface bounded with 5 faces. We will investigate each face one by one.
Face 1 : This face that lies on the y-z plane defined by the curve of the cylinder and y = 18, and z = x = 0.
- The Area of this face can be calculated by using integral calculus.
[tex]A1 = \int\limits^a_b ({1.5*y)^\frac{2}{3} } \, dy[/tex]
- The limits a,b corresponds to a = y = 18 and b = 0. Evaluate Integral.
[tex]A1 = \frac{1.31037}{1.6667} *(y^\frac{5}{3} )\limits _0^1^8 \\\\A1 = (0.786222*18^\frac{5}{3} ) = 97.19994[/tex]
Face 2: This face has the same area as Face 1. This lies on the plane x = 3 defined by the curve and y = 18, z = 0.
A1 = A2 = 97.19994 units^2
Face 3 : This face lies on the plane where y = 18 parallel to x-z plane. Representing a rectangle, defined by plane y = 18 where it intersects the cylinder and z = 0 ... Denoting width of rectangle. While the length is defined by the intersection of planes (y = 18 and x = 3).
- The intersection point of y = 18 and cylinder can be determined by equating both equations and compute the value of z coordinate.
[tex]18 = \frac{2}{3}*Z^\frac{3}{2}\\\\Z = 9[/tex]
- The area of the Face 3 can then be calculated from the formula:
A3 = Length*Width
A3 = (3)*(9)
A3 = 27 units^2
Face 4 : This face lies on the plane where z = 0 parallel i.e x-y plane. Representing a rectangle, defined by plane y = 18 where it intersects the z = 0 denoting length, while the width is defined by the intersection of planes (x = 3 and z = 0).
The area of Face 4 can be calculated directly from rectangle formula:
A4 = Length*Width
A4 = (3)*(18)
A4 = 54 units^2
Face 5 : This face is curved in nature defined by the cylinder. Which starts from plane x = 0 and limits at x = 3. We will determine the Area of this face by first computing the arc length (L) of the cylinder curve defined by:
[tex]Y = \frac{2}{3}*Z^\frac{3}{2}[/tex]
- The arc length formula (L) is given to be:
[tex]L = \int\limits^a_b {} \, ds\\\\ds = \sqrt{1 + (\frac{dy}{dz})^2 }} dz\\\\\frac{dy}{dz} = \sqrt{z} \\\\ds = \sqrt{1 + (\sqrt{z})^2 }} dz\\\\ds = \sqrt{1 +z} *dz \\\\L = \int\limits^a_b { \sqrt{1 +z}} \, dz\\\\L = \frac{2}{3}*(1+z)^(^1^.^5^) | \limits_0^9\\\\L = \frac{2}{3}*[(10.302)^(^1^.^5^) - 1 ] \\\\L = 21.37736[/tex]
- We will use the arc length and integrate with respect to x. From x = 0 to x = 3. We have:
A5 = L*( 3 - 0 )
A5 = (21.37736)*(3)
A5 = 64.13208 units^2
- The total surface area of the enclosed body is:
A = A1 + A2 + A3 + A4 + A5
A = 2*(97.1994) + 27 + 54 +64.13208
A = 339.531 units^2
