It is bounded below by the plane z 0 and figure 5 3

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Unformatted text preview: moved). Figure 4: Solution: This happens to be a Type I region. Viewed from above, the entire domain lies within the quarter circle shown in Figure 5. It is bounded below by the plane z = 0 and Figure 5: 3 above by the plane z = y , so we can set the integral up as ˆ xz dV = D ˆ 0 1ˆ √ 1 − x2 0 ˆ y xz dzdydx. (2) 0 We’ll leave the evaluation of this expression as an exercise. Cylindrical Coordinates Looking at the previous example, it might occur to you that the exterior double integral could be more conveniently expressed in polar coordinates. In fact this is exactly what we’ll do, but it’s traditional to deal with the triple integral as a whole, and for this we need a threedimensional change of coordinates. The cylindrical coordinate system is a trivial extension of the polar coordinate system; a point in R3 can be described as in Figure 6 . The transformation Figure 6: equations are simply these: x = r cos θ y = r sin θ (3) z=z The form of the Jacobian is easily extended to three variables, and the result for this particular case will not be a surprise: ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿ xr xθ xz ￿ ￿ cos θ −r sin θ 0 ￿ ￿ ￿￿ ￿ ￿￿ ￿ ∂ (x, y, z ) ￿ ￿ y y y ￿ = ￿ sin θ r cos θ 0 ￿ = r. =￿ r z￿ θ ￿ ￿ ∂ (r, θ, z ) ￿ ￿￿ ￿ ￿ ￿￿ ￿ ￿ zr zθ z z ￿ ￿ 0 0 1￿ Therefore, when switching from Cartesian to cylindrical coordinates, we can always replace dz dy dx with r dz dr dθ (and in those cases where we wish to use cylindrical coordinates, 4 this will almost always be the desired order of the differentials - you’ll almost never want to write r dr dθ dz or any of the other possibilities). Example 3: Completing the example we started above, we can calculate ˆ xz dV = D ˆ 0 = π 2 ˆ = 0 ˆ π 2 = 1 π 2 ˆ 0 = 1 2 ￿ (r cos θ)(z )(rdzdrdθ) 1 ˆ r sin θ ￿z =r sin θ ￿ r cos θ￿ drdθ ￿ 2 z =0 2z ˆ 0 ˆ r2 z cos θdzdrdθ 0 0 1 = 2 xz dzdydx 0 ˆ π 2 y 0 1 ˆ r sin θ 0 0 = ˆ ˆ 0 ˆ 1 − x2 0 ˆ 0 √ 1ˆ π 2 0 1 2 r4 sin2 θ cos θ drdθ 2 2 sin θ cos θdθ ˆ 1 r4 dr 0 ￿ π ￿ ￿...
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