Math Problem Statement

расспиши каждый шаг решения

расспиши каждый шаг решения

Solution

The problem presented in the image involves computing the surface integral of the function x2+y2\sqrt{x^2 + y^2} over the surface σ\sigma, defined by 4z2=x2+y24z^2 = x^2 + y^2 with 0z10 \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 σ\sigma is given by the equation 4z2=x2+y24z^2 = x^2 + y^2. This represents an elliptic paraboloid. In cylindrical coordinates, this can be rewritten as: r2=4z2orr=2z,r^2 = 4z^2 \quad \text{or} \quad r = 2z, where r=x2+y2r = \sqrt{x^2 + y^2} is the radial coordinate.

Step 2: Setting Up the Surface Integral

The general surface integral is: σf(x,y,z)ds.\iint_{\sigma} f(x, y, z) \, ds. For the given problem, we need to compute: σx2+y2ds,\iint_{\sigma} \sqrt{x^2 + y^2} \, ds, where dsds is the element of surface area. In cylindrical coordinates, dsds can be expressed as: ds=1+(zr)2rdθdr.ds = \sqrt{1 + \left( \frac{\partial z}{\partial r} \right)^2} \, r \, d\theta \, dr. From the surface equation r=2zr = 2z, we find: z=r2,zr=12.z = \frac{r}{2}, \quad \frac{\partial z}{\partial r} = \frac{1}{2}. Thus: ds=1+(12)2rdθdr=1+14rdθdr=52rdθdr.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 x2+y2\sqrt{x^2 + y^2} becomes rr in cylindrical coordinates, so the surface integral is: σrds=5202π02r2drdθ.\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: 02r2dr=[r33]02=83.\int_0^2 r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^2 = \frac{8}{3}. Now, the full surface integral becomes: 528302πdθ=52832π=16π53.\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: 16π53.\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 dsds 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 x2+y2x^2 + y^2 instead of x2+y2\sqrt{x^2 + y^2}?

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

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Math Problem Analysis

Mathematical Concepts

Surface Integrals
Cylindrical Coordinates
Elliptic Paraboloids

Formulas

r^2 = x^2 + y^2
ds = sqrt(1 + (∂z/∂r)^2) * r dθ dr
∫∫_σ sqrt(x^2 + y^2) ds

Theorems

Surface Integral in Cylindrical Coordinates

Suitable Grade Level

Undergraduate level (Calculus III)