Triple Integrals in Cartesian Coordinates: Setup, Volume & Mass

We start by integrating the innermost integral with respect to z, treating x and y as constants. ∫₀³(2xyz) dz = 2xy * ∫₀³z dz = 2xy * [ (1/2) z² ] from 0 to 3 = 2xy * (1/2)(9) = 2xy * (9/2) = 9xy. Next, we integrate this result with respect to y from 0 to 2: ∫₀²(9xy) dy = 9x * ∫₀²y dy = 9x * [ (1/2) y² ] from 0 to 2 = 9x * (1/2)(4) = 9x * 2 = 18x. Finally, we integrate with respect to x from 0 to 1: ∫₀¹ (18x) dx = 18 * ∫₀¹ x dx = 18 * [ (1/2) x² ] from 0 to 1 = 18 * (1/2)(1) = 18 * (1/2) = 9. Therefore, the value of the triple integral is 9.

The order of integration does not matter for volume as long as limits describe the same region.

The plane x + 2y + 3z = 6 intersects the coordinate axes at (6,0,0), (0,3,0), and (0,0,2). To set up the volume integral in the order dz dy dx, we first determine the limits for z. Solving the plane equation for z gives z = (6 - x - 2y)/3. Since the solid is bounded by z=0, we have 0 ≤ z ≤ (6 - x - 2y)/3. Next, for a fixed x, the projection in the xy-plane is the triangle bounded by x=0, y=0, and the line x + 2y = 6 (set z=0). Solving for y gives y = (6 - x)/2. So y ranges from 0 to (6 - x)/2 = 3 - (1/2)x. Finally, x ranges from 0 to 6. Therefore, the volume is ∫x=0 ⁶ ∫y=03 - (1/2)x ∫z=0(6 - x - 2y)/3 (1) dz dy dx.

We integrate step by step starting with the innermost integral with respect to z. ∫₀2 (z) dz = [ (1/2) z² ]₀2 = (1/2)(4) = 2. This result is independent of x and y, so the next integration with respect to y is ∫₀¹ (2) dy = 2 * [ y ] from 0 to 1 = 2 * 1 = 2. Finally, integrate with respect to x: ∫₀¹ (2) dx = 2 * [ x ] from 0 to 1 = 2 * 1 = 2. Therefore, the value of the triple integral is 2.

Mass is computed as the triple integral of density over the region. So mass = ∫₀¹ ∫₀¹ ∫₀¹ (x + y + z) dz dy dx. First, integrate with respect to z: ∫₀¹ (x + y + z) dz = [ (x+y)z + (1/2)z² ] from 0 to 1 = (x+y)(1) + (1/2)(1) = x + y + 1/2. Next, integrate this with respect to y: ∫₀¹ (x + y + 1/2) dy = [ x y + (1/2) y² + (1/2) y ] from 0 to 1 = x(1) + (1/2)(1) + (1/2)(1) = x + 1/2 + 1/2 = x + 1. Finally, integrate with respect to x: ∫₀¹ (x + 1) dx = [ (1/2) x² + x ] from 0 to 1 = (1/2)(1) + 1 = 1/2 + 1 = 3/2. Thus, the mass is 3/2.

In cylindrical coordinates, x = r cos θ, y = r sin θ, so x² + y² = r². The volume element dV becomes r dz dr dθ. The density is constant ρ = 5. The moment of inertia about the z-axis is ∭ (x² + y²) ρ dV = ∭ (r²) * 5 * (r dz dr dθ) = ∭ 5 r³ dz dr dθ. The region is: r from 0 to 2 (since x²+y² ≤ 4 implies r ≤ 2), θ from 0 to 2π, and z from 0 to 3. So the integral becomes ∫ from θ=0 to 2π ∫ from r=0 to 2 ∫ from z=0 to 3 (5 r³) dz dr dθ. This matches option A because (r²) * 5 * r = 5 r³. Option B misses the extra r from dV. Option C uses (r) instead of (r²). Option D integrates density only, not the moment of inertia.

The original region is defined by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1-x, and 0 ≤ z ≤ 1-x-y. This is a tetrahedron bounded by the planes x=0, y=0, z=0, and x+y+z=1. To change the order to dx dy dz, we need to describe the region as: for fixed z, x and y vary such that x ≥ 0, y ≥ 0, and x+y ≤ 1-z. So for fixed z, y ranges from 0 to 1-z, and for fixed y and z, x ranges from 0 to 1 - y - z. Also, z ranges from 0 to 1. Therefore, the integral becomes ∫₀¹ ∫₀¹-z ∫₀¹-y-z f(x,y,z) dx dy dz. This matches option A. Option B has the order dx dz dy, not dx dy dz. Option C integrates over a cube. Option D has incorrect limits for x (it should be 1-y-z, not 1-y).

In physics and engineering, triple integrals are used to compute total quantities over a three-dimensional region. If f(x,y,z) represents density (mass per unit volume), then the triple integral gives total mass. If f(x,y,z) represents charge density (charge per unit volume), then the triple integral gives total electric charge. If f(x,y,z) is a probability density function for a continuous random vector (X,Y,Z), then the triple integral over a region gives the probability that (X,Y,Z) lies in that region. Therefore, all the listed options are valid interpretations.

We integrate step by step. First, integrate with respect to z: ∫₀²(y sin(x)) dz = y sin(x) * [ z ] from 0 to 2 = y sin(x) * 2 = 2y sin(x). Next, integrate with respect to y: ∫₀¹ (2y sin(x)) dy = 2 sin(x) * ∫₀¹ y dy = 2 sin(x) * [ (1/2) y² ] from 0 to 1 = 2 sin(x) * (1/2) = sin(x). Finally, integrate with respect to x: ∫ from 0 to π/2 sin(x) dx = [ -cos(x) ] from 0 to π/2 = -cos(π/2) - (-cos(0)) = -0 + 1 = 1. Thus, the value is 1.

The solid is bounded below by z=0 and above by the paraboloid z = 4 - x² - y². The projection onto the xy-plane is the disk x² + y² ≤ 4. In Cartesian coordinates, the volume is ∫∫ over the disk of (4 - x² - y²) dA. Setting up the triple integral in the order dz dy dx: for fixed (x,y), z goes from 0 to 4 - x² - y²; then y goes from -√(4-x²) to √(4-x²); and x goes from -2 to 2. So option A is correct. In cylindrical coordinates, x = r cos θ, y = r sin θ, the paraboloid becomes z = 4 - r², and the disk is 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π. The volume element is r dz dr dθ, and z goes from 0 to 4 - r². So the integral is ∫ from 0 to 2π ∫₀²∫ from 0 to 4-r² (r) dz dr dθ, which is option B. Both integrals represent the same volume, so C is correct.

First, compute the volume of the cube: 1*1*1 = 1. Next, compute the triple integral of f over the cube: ∫₀¹ ∫₀¹ ∫₀¹ (xyz) dz dy dx. We can compute each integral separately:∫₀¹ x dx = 1/2,∫₀¹ y dy = 1/2,∫₀¹ z dz = 1/2. So the product is (1/2)*(1/2)*(1/2) = 1/8. Therefore, the average value is (1/8) / 1 = 1/8.

The z-coordinate of the center of mass for a solid with constant density is given by the formula z-bar = (1/Volume) times the triple integral of z over the region. For the tetrahedron bounded by the coordinate planes and the plane x + y + z = 1, the region of integration is defined by 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 - x, and 0 ≤ z ≤ 1 - x - y.  First, compute the volume V. Set up the triple integral: V = ∫ from 0 to 1 ∫ from 0 to 1-x ∫ from 0 to 1-x-y 1 dz dy dx. The innermost integral evaluates to 1 - x - y. Then V = ∫ from 0 to 1 ∫ from 0 to 1-x (1 - x - y) dy dx. Integrate with respect to y: the antiderivative is (1-x)y - y²/2, evaluated from 0 to 1-x gives (1-x)² - (1-x)²/2 = (1-x)²/2. So V = ∫ from 0 to 1 (1-x)²/2 dx. Let u = 1-x, then du = -dx, and when x=0, u=1; when x=1, u=0. Thus ∫ from 0 to 1 (1-x)² dx = ∫ from 1 to 0 u² (-du) = ∫ from 0 to 1 u² du = u³/3 evaluated from 0 to 1 = 1/3. Therefore, V = (1/2)*(1/3) = 1/6.

Next, compute the triple integral of z over the region: ∭ z dV = ∫ from 0 to 1 ∫ from 0 to 1-x ∫ from 0 to 1-x-y z dz dy dx. The innermost integral ∫ z dz from 0 to 1-x-y gives (1-x-y)²/2. So ∭ z dV = ∫ from 0 to 1 ∫ from 0 to 1-x (1-x-y)²/2 dy dx. Compute the inner integral with respect to y: ∫ from 0 to 1-x (1-x-y)² dy. Let u = 1-x-y, then dy = -du, and when y=0, u=1-x; when y=1-x, u=0. So the integral becomes ∫ from 1-x to 0 u² (-du) = ∫ from 0 to 1-x u² du = u³/3 evaluated from 0 to 1-x = (1-x)³/3. Thus ∭ z dV = ∫ from 0 to 1 (1/2)*(1-x)³/3 dx = ∫ from 0 to 1 (1-x)³/6 dx. Again, let u = 1-x, then dx = -du, and the integral becomes ∫ from 1 to 0 u³/6 (-du) = ∫ from 0 to 1 u³/6 du = (1/6)* (u⁴/4) evaluated from 0 to 1 = (1/6)*(1/4) = 1/24. Finally, z-bar = (1/24) / (1/6) = (1/24) * 6 = 6/24 = 1/4

The moment of inertia about the x-axis is I_x = ∭ (y² + z²) dV. Since the density is 1, we compute ∫₀¹∫₀¹∫₀¹ (y² + z²) dx dy dz. We can integrate in any order. First, note that the integral is separable:∫₀¹ dx = 1, so I_x =∫₀¹ ∫₀¹ (y² + z²) dy dz. Compute∫₀¹ ∫₀¹ y² dy dz =∫₀¹ [ (⅓) y³ ]₀¹ dz =∫₀¹ (⅓) dz = 1/3. Similarly,∫₀¹ ∫₀¹ z² dy dz =∫₀¹ z² (∫₀¹ dy ) dz =∫₀¹ z² * 1 dz = [ (⅓) z³ ]₀¹ = 1/3. So I_x = 1/3 + 1/3 = 2/3.

We start by visualizing the region. The solid is bounded below by the paraboloid z = x² + y² and above by the plane z = 4. In cylindrical coordinates, x² + y² becomes r², so the lower bound for z is r² and the upper bound is 4. This establishes the z-limits as r² to 4. Next, we determine the shadow of the region on the xy-plane (where z=0). The intersection of the paraboloid and the plane occurs where r² = 4, which means r = 2. Therefore, the radius r goes from 0 to 2. Since the paraboloid is rotationally symmetric around the z-axis and encompasses the full circle, the angle theta ranges from 0 to 2π.

When we transform coordinates in multiple integrals, we must include the Jacobian of the coordinate transformation to account for the distortion of the volume element. For cylindrical coordinates, where x = r cos(theta) and y = r sin(theta), the Jacobian determinant calculation results in a factor of r. Therefore, the rectangular volume element dx dy dz is replaced by r dz dr dtheta (or any permutation of these differentials). If we omit the r, we are not accurately representing the physical volume of the wedge-shaped elements in the cylindrical grid.