Question

The surface area of the plane \(x + 2y + 2z = 12\) cut off by \(x=0, y=0\) and \(x^2+y^2=16\) is

• A. 2\(\pi\)
• B. 3\(\pi\)
• C. 5\(\pi\)
• D. 6\(\pi\)

Hint:

The formula \( S = \frac{A_{proj}}{|\cos\theta|} \), where \(A_{proj}\) is the area of the projection and \(\theta\) is the angle between the normal to the surface and the normal to the projection plane, is very powerful for finding the area of a planar region. Here, \(\cos\theta = \frac{\vec{n} \cdot \hat{k}}{|\vec{n}||\hat{k}|} = \frac{2}{3}\), so \(S = \frac{4\pi}{2/3} = 6\pi\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need to find the area of a portion of a plane. The portion is defined by its projection onto the xy-plane. The projection is bounded by \(x=0, y=0\) and \(x^2+y^2=16\), which means it's the part of the circle \(x^2+y^2=16\) that lies in the first quadrant.
Step 2: Key Formula or Approach:
The formula for the surface area (S) of a surface \(z = f(x, y)\) over a region R in the xy-plane is: \[ S = \iint_R \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dA \] This can be simplified. If the normal to the plane is \( \vec{n} \) and the projection is onto the xy-plane, the surface area is \( S = \frac{A_{xy}}{|\vec{n} \cdot \hat{k}|} \), where \(A_{xy}\) is the area of the projection.
Step 3: Detailed Explanation:
Method 1: Using the integral formula
From the plane equation \(x + 2y + 2z = 12\), we can express z as: \[ z = 6 - \frac{1}{2}x - y \] Now, find the partial derivatives: \[ \frac{\partial z}{\partial x} = -\frac{1}{2} \] \[ \frac{\partial z}{\partial y} = -1 \] The integrand for the surface area is: \[ \sqrt{1 + \left(-\frac{1}{2}\right)^2 + (-1)^2} = \sqrt{1 + \frac{1}{4} + 1} = \sqrt{\frac{4+1+4}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2} \] The surface area is: \[ S = \iint_R \frac{3}{2} \, dA = \frac{3}{2} \times (\text{Area of R}) \] The region R is the part of the circle \(x^2+y^2=16\) in the first quadrant (\(x \ge 0, y \ge 0\)). The radius of the circle is \(r = \sqrt{16} = 4\). The area of the full circle is \( \pi r^2 = \pi (4)^2 = 16\pi \). The area of the region R (the first quadrant) is one-fourth of the total area: \[ \text{Area of R} = \frac{1}{4} (16\pi) = 4\pi \] Now, calculate the surface area S: \[ S = \frac{3}{2} \times (4\pi) = 6\pi \] Method 2: Using the projection formula
The normal vector to the plane \(x + 2y + 2z = 12\) is \( \vec{n} = \langle 1, 2, 2 \rangle \). The unit normal to the xy-plane is \( \hat{k} = \langle 0, 0, 1 \rangle \). The area of the projection \(A_{xy}\) is \(4\pi\). \[ |\vec{n} \cdot \hat{k}| = |\langle 1, 2, 2 \rangle \cdot \langle 0, 0, 1 \rangle| = |2| = 2 \] The magnitude of the normal vector is \( |\vec{n}| = \sqrt{1^2+2^2+2^2} = \sqrt{9} = 3 \). The formula relating surface area S and projected area \(A_{xy}\) is \( S = \frac{|\vec{n}|}{|\vec{n} \cdot \hat{k}|} A_{xy} \). \[ S = \frac{3}{2} \times (4\pi) = 6\pi \] Step 4: Final Answer:
The surface area of the specified portion of the plane is \(6\pi\).