Question

For each t ∈ (0, 1), the surface P_t in \(\R^3\) is defined by
\(P_t = \left\{(x, y, z) : (x^2 + y^2 )z = 1, t^2 ≤ x^2 + y^2 ≤ 1\right\}.\)
Let a_t ∈ R be the surface area of P_t. Then

• A. \(a_t=\iint\limits_{t^2\le x^2 + y^2\le 1}\sqrt{1+\frac{4x^2}{(x^2+y^2)^4}+\frac{4y^2}{(x^2+y^2)^4}}dx\ dy\)
• B. \(a_t=\iint\limits_{t^2\le x^2 + y^2\le 1}\sqrt{1+\frac{4x^2}{(x^2+y^2)^2}+\frac{4y^2}{(x^2+y^2)^2}}dx\ dy\)
• C. the limit\(\lim\limits_ {t→0^+}a_t\) does NOT exist
• D. the limit\(\lim\limits_ {t→0^+}a_t\) exist

Answer 1

The Correct Option is A, C

To solve this problem, we need to determine the surface area of the surface \(P_t\) defined by the given conditions and analyze the limit behavior as \(t \to 0^+\).

The surface \(P_t\) is given by:

\( P_t = \left\{(x, y, z) : (x^2 + y^2)z = 1, \, t^2 \le x^2 + y^2 \le 1 \right\}\)

We identify \(z\) in terms of \(x\) and \(y\):

\(z = \frac{1}{x^2 + y^2}\)

The expression for the area element \(dS\) on this surface can be found using the formula:

\(dS = \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} \, dx \, dy\)

Calculating the partial derivatives:

• \(\frac{\partial z}{\partial x} = \frac{-2x}{(x^2 + y^2)^2}\)
• \(\frac{\partial z}{\partial y} = \frac{-2y}{(x^2 + y^2)^2}\)

Substituting these into the formula for \(dS\), we get:

\(dS = \sqrt{1 + \left(\frac{-2x}{(x^2+y^2)^2}\right)^2 + \left(\frac{-2y}{(x^2+y^2)^2}\right)^2}\, dx \, dy\)

Simplifying,

\(dS = \sqrt{1 + \frac{4x^2}{(x^2 + y^2)^4} + \frac{4y^2}{(x^2 + y^2)^4}} \, dx \, dy\)

The surface area \(a_t\) is thus given by the integral:

\(a_t = \iint\limits_{t^2 \le x^2 + y^2 \le 1}\sqrt{1+\frac{4x^2}{(x^2+y^2)^4}+\frac{4y^2}{(x^2+y^2)^4}} \, dx \, dy\)

This confirms the first part of the correct answer.

Now, we analyze the behavior of \(a_t\) as \(t \to 0^+\). As \(t\) approaches zero, the region \(t^2 \le x^2 + y^2 \le 1\) approaches the annular region that almost covers the entire disk \(x^2 + y^2 \le 1\) except the origin.

However, near the origin, the function \(\sqrt{1+\frac{4x^2}{(x^2+y^2)^4}+\frac{4y^2}{(x^2+y^2)^4}}\) becomes singular due to the denominator \((x^2+y^2)^4\) becoming very small, making the limit of the integral problematic.

Hence, the limit \(\lim\limits_{t \to 0^+} a_t\) does not exist, confirming the second part of the correct answer.