Math Problem Statement

volume of a solid generated by a surface z= 16-x2-y2 over the region bounded by y=√2x, z axis and y=4x-2

Solution

To find the volume of the solid generated by the surface z=16x2y2z = 16 - x^2 - y^2 over the given region, we need to carefully set up and evaluate a double integral over the specified region. Here's the step-by-step solution:

1. Understand the Region of Integration

The region is bounded by:

  • y=2xy = \sqrt{2}x (line 1),
  • y=4x2y = 4x - 2 (line 2),
  • xx-axis (y0y \geq 0),
  • yy-axis (x0x \geq 0).

The intersection of y=2xy = \sqrt{2}x and y=4x2y = 4x - 2 gives the bounds in xx and yy:

  1. Set 2x=4x2\sqrt{2}x = 4x - 2: 2x4x=2    x(24)=2    x=242.\sqrt{2}x - 4x = -2 \implies x(\sqrt{2} - 4) = -2 \implies x = \frac{2}{4 - \sqrt{2}}.

    Simplify the fraction: x=2(4+2)162=2(4+2)14=2+27.x = \frac{2(4 + \sqrt{2})}{16 - 2} = \frac{2(4 + \sqrt{2})}{14} = \frac{2 + \sqrt{2}}{7}.

  2. Use this xx-value to find yy-coordinates: Substitute x=2+27x = \frac{2 + \sqrt{2}}{7} into either equation y=2xy = \sqrt{2}x: y=22+27=22+27.y = \sqrt{2} \cdot \frac{2 + \sqrt{2}}{7} = \frac{2\sqrt{2} + 2}{7}.

2. Volume Integral

The volume is given by: V=R(16x2y2)dA,V = \int \int_R (16 - x^2 - y^2) \, dA, where RR is the region bounded by the lines y=2xy = \sqrt{2}x and y=4x2y = 4x - 2.

Setting Bounds:

  1. xx-bounds are 00 to 2+27\frac{2 + \sqrt{2}}{7},
  2. For a fixed xx, yy-bounds are from y=2xy = \sqrt{2}x to y=4x2y = 4x - 2.

Thus, the integral becomes: V=02+272x4x2(16x2y2)dydx.V = \int_0^{\frac{2 + \sqrt{2}}{7}} \int_{\sqrt{2}x}^{4x - 2} (16 - x^2 - y^2) \, dy \, dx.

3. Solve the Inner Integral

For the inner integral: 2x4x2(16x2y2)dy=2x4x216dy2x4x2x2dy2x4x2y2dy.\int_{\sqrt{2}x}^{4x - 2} (16 - x^2 - y^2) \, dy = \int_{\sqrt{2}x}^{4x - 2} 16 \, dy - \int_{\sqrt{2}x}^{4x - 2} x^2 \, dy - \int_{\sqrt{2}x}^{4x - 2} y^2 \, dy.

  1. 2x4x216dy=16[y]2x4x2=16((4x2)(2x))\int_{\sqrt{2}x}^{4x - 2} 16 \, dy = 16[y]_{\sqrt{2}x}^{4x - 2} = 16((4x - 2) - (\sqrt{2}x)).
  2. 2x4x2x2dy=x2[y]2x4x2=x2((4x2)(2x))\int_{\sqrt{2}x}^{4x - 2} x^2 \, dy = x^2[y]_{\sqrt{2}x}^{4x - 2} = x^2((4x - 2) - (\sqrt{2}x)).
  3. 2x4x2y2dy=[y33]2x4x2=(4x2)33(2x)33\int_{\sqrt{2}x}^{4x - 2} y^2 \, dy = \left[\frac{y^3}{3}\right]_{\sqrt{2}x}^{4x - 2} = \frac{(4x - 2)^3}{3} - \frac{(\sqrt{2}x)^3}{3}.

4. Evaluate the Outer Integral

Substitute the results of the inner integrals into the outer integral and simplify. This involves integrating with respect to xx over [0,2+27][0, \frac{2 + \sqrt{2}}{7}].

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Math Problem Analysis

Mathematical Concepts

Volume of solids
Double integrals
Region of integration
Algebraic manipulation

Formulas

Volume = ∬_R (surface function) dA
Intersection of two lines to find bounds

Theorems

Fundamental theorem of calculus for double integrals
Basic properties of integration

Suitable Grade Level

Undergraduate (Calculus II or III)