请解答一下这道题目---**Question 3** **Question Stem:** Find the volume of the solid bounded by z = 6 - x^2 - y^2 and z = sqrt(x^2 + y^2) by using triple integrals.

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我们需要求由抛物面 z = 6 - x² - y² 和圆锥面 z = √(x² + y²) 围成的立体体积。在 y = 0 截面上，抛物面表现为开口向下的抛物线，圆锥面表现为直线。两曲面相交于 r = 2 处，形成一个有界区域。 We need to find the volume of a solid bounded by two surfaces: z = 6 - x² - y², which is a paraboloid opening downward, and z = √(x² + y²), which is a cone. We'll solve this using triple integrals. To find the intersection, we set the two surfaces equal: 6 - x² - y² = √(x² + y²). Converting to cylindrical coordinates where x² + y² = r², we get 6 - r² = r. Rearranging gives r² + r - 6 = 0, which factors as (r + 3)(r - 2) = 0. Since r ≥ 0, we have r = 2. This means the projection region is a disk of radius 2. To find the volume, we integrate the difference between the upper and lower surfaces. The upper surface is z = 6 - r² and the lower surface is z = r. In cylindrical coordinates, our triple integral becomes the integral from 0 to 2π for θ, from 0 to 2 for r, and from r to 6 - r² for z, with the volume element r dz dr dθ. Now we evaluate the integral step by step. First, we integrate with respect to z from r to 6 - r², giving us r times (6 - r² - r). Next, we integrate with respect to r from 0 to 2, expanding to get the integral of 6r - r² - r³. Evaluating this gives us 3r² - r³/3 - r⁴/4 from 0 to 2, which equals 12 - 8/3 - 4, simplifying to 16/3. Finally, we integrate with respect to θ from 0 to 2π. Since our integrand 16/3 is constant with respect to θ, we get 16/3 times 2π, which equals 32π/3. Therefore, the volume of the solid bounded by the two surfaces is 32π/3 cubic units. 现在设置三重积分。体积等于上表面减去下表面在区域D上的积分，其中D是圆盘x²+y²≤4。转换为柱坐标后，积分变为从0到2π对θ积分，从0到2对r积分，从r到6-r²对z积分。右图显示了一个柱坐标微元，底面是极坐标网格小扇形，高度从圆锥面延伸到抛物面。 We need to find the volume of the solid bounded by two surfaces: a paraboloid z equals 6 minus x squared minus y squared, and a cone z equals square root of x squared plus y squared. We will use triple integrals to solve this problem. First, we find where the two surfaces intersect. Setting 6 minus x squared minus y squared equal to square root of x squared plus y squared, we convert to cylindrical coordinates. This gives us 6 minus r squared equals r, which simplifies to r squared plus r minus 6 equals zero. Factoring, we get r equals 2 as the positive solution. Our integration region is 0 to 2 for r and 0 to 2π for theta. Now we set up the triple integral in cylindrical coordinates. The volume is the integral from 0 to 2π for theta, 0 to 2 for r, and from r to 6 minus r squared for z, with integrand r times dz dr d-theta. The factor r comes from the Jacobian determinant in cylindrical coordinates. The z limits are from the cone surface at z equals r to the paraboloid surface at z equals 6 minus r squared. Now we calculate the integral step by step. First, integrate with respect to z from r to 6 minus r squared, getting r times the difference, which is r times 6 minus r minus r squared. Next, integrate with respect to r from 0 to 2, expanding to get the integral of 6r minus r squared minus r cubed. Evaluating gives us 3r squared minus r cubed over 3 minus r to the fourth over 4, which equals 12 minus 8 thirds minus 4, or 16 thirds. Finally, integrate with respect to theta from 0 to 2π, giving us 2π times 16 thirds, which equals 32π over 3. Therefore, the volume of the solid bounded by the paraboloid z equals 6 minus x squared minus y squared and the cone z equals square root of x squared plus y squared is 32π over 3 cubic units. This completes our solution using triple integrals in cylindrical coordinates. 最后一步是对θ从0到2π积分。由于被积函数16/3是常数，结果为16/3乘以2π，等于32π/3。因此，由抛物面z=6-x²-y²和圆锥面z=√(x²+y²)围成的立体体积为32π/3立方单位。我们可以验证：当r=0时高度为6，当r=2时两曲面相交，体积为正值符合几何直观。