Math Problem Statement

Find the area of the surface. the part of the surface z = xy that lies within the cylinder x2 + y2 = 64

Solution

To find the area of the surface z=xyz = xy that lies within the cylinder x2+y2=64x^2 + y^2 = 64, we can use surface area formulas from multivariable calculus.

Steps:

  1. Surface Area Formula: The general formula for the surface area AA of a parametric surface z=f(x,y)z = f(x, y) is: A=D1+(zx)2+(zy)2dAA = \iint_{D} \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2 } \, dA where DD is the projection of the surface onto the xyxy-plane.

  2. Partial Derivatives: Since z=xyz = xy, we first compute the partial derivatives: zx=y,zy=x\frac{\partial z}{\partial x} = y, \quad \frac{\partial z}{\partial y} = x

    The square of these derivatives are: (zx)2=y2,(zy)2=x2\left( \frac{\partial z}{\partial x} \right)^2 = y^2, \quad \left( \frac{\partial z}{\partial y} \right)^2 = x^2

  3. Surface Area Integral: The surface area formula becomes: A=D1+x2+y2dAA = \iint_{D} \sqrt{1 + x^2 + y^2} \, dA where DD is the disk defined by x2+y264x^2 + y^2 \leq 64.

  4. Convert to Polar Coordinates: To simplify the integral, we convert to polar coordinates where x=rcosθx = r\cos\theta and y=rsinθy = r\sin\theta. The area element dAdA in polar coordinates is rdrdθr \, dr \, d\theta, and the equation x2+y2=r2x^2 + y^2 = r^2.

    So the integral becomes: A=02π081+r2rdrdθA = \int_0^{2\pi} \int_0^8 \sqrt{1 + r^2} \, r \, dr \, d\theta

  5. Evaluate the Radial Integral: First, compute the radial part of the integral: 081+r2rdr\int_0^8 \sqrt{1 + r^2} \, r \, dr

    Let u=1+r2u = 1 + r^2, so du=2rdrdu = 2r \, dr. This transforms the integral into: 12165udu=12[23u3/2]165=13(653/21)\frac{1}{2} \int_1^{65} \sqrt{u} \, du = \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_1^{65} = \frac{1}{3} \left( 65^{3/2} - 1 \right)

    Now calculate 653/265^{3/2}: 653/2=65×6565×8.06=523.965^{3/2} = 65 \times \sqrt{65} \approx 65 \times 8.06 = 523.9

    So the radial integral evaluates to: 13(523.91)=13×522.9174.3\frac{1}{3} \left( 523.9 - 1 \right) = \frac{1}{3} \times 522.9 \approx 174.3

  6. Complete the Surface Area Integral: Multiply by the angular part, which is 02πdθ=2π\int_0^{2\pi} d\theta = 2\pi: A=2π×174.3=348.6πA = 2\pi \times 174.3 = 348.6\pi

Final Answer:

The surface area of the part of the surface z=xyz = xy that lies within the cylinder x2+y2=64x^2 + y^2 = 64 is approximately: A348.6πsquare unitsA \approx 348.6\pi \quad \text{square units}

Would you like more details or have any questions?

Here are 5 related questions to consider:

  1. How do polar coordinates simplify surface area integrals?
  2. What is the geometric interpretation of the surface z=xyz = xy?
  3. How do we compute surface area for surfaces in different coordinate systems?
  4. Why do we use the cylinder x2+y2=64x^2 + y^2 = 64 as the boundary in this problem?
  5. How do changes in surface height zz affect the total surface area?

Tip: Converting to polar coordinates is a powerful technique for symmetry-based problems, especially when the boundary is circular.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Surface Area
Polar Coordinates

Formulas

Surface area formula: A = ∬_D √(1 + (∂z/∂x)^2 + (∂z/∂y)^2) dA
Partial derivatives of z = xy: ∂z/∂x = y, ∂z/∂y = x
Surface area integral in polar coordinates: A = ∫_0^2π ∫_0^8 √(1 + r^2) r dr dθ

Theorems

Surface area of parametric surfaces
Conversion to polar coordinates for integrals

Suitable Grade Level

University level (Calculus III)