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Chapter 9: Problem 38

Describe the solid satisfying the condition. \(x^{2}+y^{2}+z^{2}>4\)

Short Answer

Expert verified
The solid is the region in 3D space outside a sphere of radius 2 centered at the origin.

Step by step solution

01

Identification of the Equation Form

The equation provided, \(x^{2}+y^{2}+z^{2}>4\), is of the form \(x^{2}+y^{2}+z^{2}=r^{2}\), where r is the radius of a sphere in a 3D space. Reading from the equation, we can see that r is equal to 2 (since 2 squared equals 4). As we are dealing with an inequality and not an equation, the radius is not exactly 2, but the sphere's expansion will be based on this value of 2.
02

Interpretation of the Inequality

Since we have \(x^{2}+y^{2}+z^{2}>4\), this conveys that we are dealing with a volume that is greater than the volume of the sphere with radius 2. This simply means that the solid that satisfies this condition is the region of the 3D space located outside of a sphere of radius 2, centered at the origin.
03

Final Conclusion

Having identified the inequality's nature and the type of solid it defines, we can therefore define the solid that satisfies the condition \(x^{2}+y^{2}+z^{2}>4\) as the region outside a sphere of radius 2 centered at the origin in a three-dimensional (3D) coordinate space.

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Let \(\mathbf{u}=\mathbf{i}+\mathbf{j}, \mathbf{v}=\mathbf{j}+\mathbf{k},\) and \(\mathbf{w}=a \mathbf{u}+b \mathbf{v} .\) (a) Sketch \(\mathbf{u}\) and \(\mathbf{v}\). (b) If \(\mathbf{w}=\mathbf{0}\), show that \(a\) and \(b\) must both be zero. (c) Find \(a\) and \(b\) such that \(\mathbf{w}=\mathbf{i}+2 \mathbf{j}+\mathbf{k}\). (d) Show that no choice of \(a\) and \(b\) yields \(\mathbf{w}=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}\).
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