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The problem presented in the image involves computing the surface integral of the function \( \sqrt{x^2 + y^2} \) over the surface \( \sigma \), defined by \( 4z^2 = x^2 + y^2 \) with \( 0 \leq z \leq 1 \).

Let's break down the steps to solve this surface integral step-by-step:

### Step 1: Understanding the Surface
The surface \( \sigma \) is given by the equation \( 4z^2 = x^2 + y^2 \). This represents an elliptic paraboloid. In cylindrical coordinates, this can be rewritten as:
\[
r^2 = 4z^2 \quad \text{or} \quad r = 2z,
\]
where \( r = \sqrt{x^2 + y^2} \) is the radial coordinate.

### Step 2: Setting Up the Surface Integral
The general surface integral is:
\[
\iint_{\sigma} f(x, y, z) \, ds.
\]
For the given problem, we need to compute:
\[
\iint_{\sigma} \sqrt{x^2 + y^2} \, ds,
\]
where \( ds \) is the element of surface area. In cylindrical coordinates, \( ds \) can be expressed as:
\[
ds = \sqrt{1 + \left( \frac{\partial z}{\partial r} \right)^2} \, r \, d\theta \, dr.
\]
From the surface equation \( r = 2z \), we find:
\[
z = \frac{r}{2}, \quad \frac{\partial z}{\partial r} = \frac{1}{2}.
\]
Thus:
\[
ds = \sqrt{1 + \left( \frac{1}{2} \right)^2} \, r \, d\theta \, dr = \sqrt{1 + \frac{1}{4}} \, r \, d\theta \, dr = \frac{\sqrt{5}}{2} \, r \, d\theta \, dr.
\]

### Step 3: Substituting into the Integral
The integrand \( \sqrt{x^2 + y^2} \) becomes \( r \) in cylindrical coordinates, so the surface integral is:
\[
\iint_{\sigma} r \, ds = \frac{\sqrt{5}}{2} \int_0^{2\pi} \int_0^2 r^2 \, dr \, d\theta.
\]

### Step 4: Solving the Integral
First, compute the radial integral:
\[
\int_0^2 r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^2 = \frac{8}{3}.
\]
Now, the full surface integral becomes:
\[
\frac{\sqrt{5}}{2} \cdot \frac{8}{3} \int_0^{2\pi} d\theta = \frac{\sqrt{5}}{2} \cdot \frac{8}{3} \cdot 2\pi = \frac{16\pi \sqrt{5}}{3}.
\]

### Final Answer:
The value of the surface integral is:
\[
\boxed{\frac{16\pi \sqrt{5}}{3}}.
\]

---

Let me know if you'd like further details or if anything is unclear!

Here are some additional questions related to this topic:

1. How does the surface element \( ds \) change for different surfaces?
2. What is the geometric interpretation of this surface integral?
3. Can this surface integral be computed in spherical coordinates?
4. How would the problem change if the surface were a cone instead of a paraboloid?
5. What if the function to integrate was \( x^2 + y^2 \) instead of \( \sqrt{x^2 + y^2} \)?

**Tip**: Cylindrical coordinates are particularly useful for surfaces with rotational symmetry around the z-axis, like cones or paraboloids.

Learn how to compute the surface integral of sqrt(x^2 + y^2) over the surface defined by 4z^2 = x^2 + y^2, using cylindrical coordinates. Detailed step-by-step solution provided, suitable for Calculus III students.

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