Mathematics-II (BCA-201) - 2017

 

Section A (Very Short Answer Questions). Each question carries 3 marks.

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1. Let A = {2, 3, 5}, B = {3, 6, 8}, C = {4, 7, 9}. Show that A × (B ∩ C) = (A × B) ∩ (A × C).

Solution:

1. Step 1: Understand the operation.

   • $A × B$: Cartesian product of sets $A$ and $B$.
   • $B ∩ C$: Intersection of sets $B$ and $C$.
   • $A × (B ∩ C)$: Cartesian product of $A$ with the intersection of $B$ and $C$.
   • $(A × B) ∩ (A × C)$: Intersection of the Cartesian products of $A × B$ and $A × C$.

2. Step 2: Compute the sets.

   • $A = \{2, 3, 5\}$, $B = \{3, 6, 8\}$, $C = \{4, 7, 9\}$.
   • $B ∩ C = \phi$ (since there are no common elements between $B$ and $C$).

3. Step 3: Compute $A × (B ∩ C)$.

   • Since $B ∩ C = \phi$, $A × (B ∩ C) = \phi$.

4. Step 4: Compute $(A × B) ∩ (A × C)$.

   • $A × B = \{(2, 3), (2, 6), (2, 8), (3, 3), (3, 6), (3, 8), (5, 3), (5, 6), (5, 8)\}$,
   • $A × C = \{(2, 4), (2, 7), (2, 9), (3, 4), (3, 7), (3, 9), (5, 4), (5, 7), (5, 9)\}$,
   • Intersection $(A × B) ∩ (A × C) = \phi$ (since there are no common pairs).

5. Conclusion:

   • Both $A × (B ∩ C) = \phi$ and $(A × B) ∩ (A × C) = \phi$,
   • Therefore, $A × (B ∩ C) = (A × B) ∩ (A × C)$.

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2. Let Q be the set of rational numbers. Let $f : Q → Q$ be defined by $f(x) = 2x + 3$. Show that $f$ is bijective.

Solution:

1. Check for injectivity (one-to-one):

   • Assume $f(x_1) = f(x_2)$,
   • $2x_1 + 3 = 2x_2 + 3$,
   • Subtract 3 from both sides: $2x_1 = 2x_2$,
   • Divide by 2: $x_1 = x_2$,
   • Thus, $f$ is injective.

2. Check for surjectivity (onto):

   • Let $y \in Q$ (an arbitrary rational number),
   • Solve $y = 2x + 3$ for $x$: $$x = \frac{y - 3}{2}.$$
   • Since $y$ and 3 are rational, and $2 \neq 0$, $x$ is rational.
   • Thus, for every $y \in Q$, there exists $x \in Q$ such that $f(x) = y$,
   • Therefore, $f$ is surjective.

3. Conclusion:

   • $f$ is both injective and surjective, hence bijective.

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3. Show that the set of all factors of 12 under divisibility forms a lattice.

Solution:

1. Factors of 12:

   • The factors of 12 are $\{1, 2, 3, 4, 6, 12\}$.

2. Partial order:

   • Define the partial order $a \leq b$ if $a$ divides $b$.

3. Lattice verification:

   • A lattice is a partially ordered set in which every two elements have:
     • A least upper bound (LUB) (also called join),
     • A greatest lower bound (GLB) (also called meet).

4. Check join and meet:

   • For any two elements $a, b$:
     • The LUB is the smallest factor of 12 divisible by both $a$ and $b$,
     • The GLB is the largest factor of 12 that divides both $a$ and $b$.
   • For example:
     • $a = 2, b = 3$: $\text{LUB} = 6, \text{GLB} = 1$,
     • $a = 4, b = 6$: $\text{LUB} = 12, \text{GLB} = 2$.

5. Conclusion:

   • The set of factors of 12 with the divisibility relation forms a lattice.

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4. If $U = f(y/x)$, show that $x \frac{\partial U}{\partial x} + y \frac{\partial U}{\partial y} = 0$.

Solution:

1. Given:

   • $U = f(y/x)$, let $v = y/x$.

2. Partial derivatives:

   • $\frac{\partial v}{\partial x} = -\frac{y}{x^2}$,
   • $\frac{\partial v}{\partial y} = \frac{1}{x}$.

3. Chain rule:

   • $\frac{\partial U}{\partial x} = \frac{df}{dv} \cdot \frac{\partial v}{\partial x} = f'(v) \cdot \left(-\frac{y}{x^2}\right)$,
   • $\frac{\partial U}{\partial y} = \frac{df}{dv} \cdot \frac{\partial v}{\partial y} = f'(v) \cdot \frac{1}{x}$.

4. Substitute into equation:

   • $x \frac{\partial U}{\partial x} + y \frac{\partial U}{\partial y}$,
   • $x \cdot f'(v) \cdot \left(-\frac{y}{x^2}\right) + y \cdot f'(v) \cdot \frac{1}{x}$,
   • $-f'(v) \cdot \frac{xy}{x^2} + f'(v) \cdot \frac{y}{x}$,
   • $-\frac{f'(v) \cdot y}{x} + \frac{f'(v) \cdot y}{x} = 0$.

5. Conclusion:

   • The given equation is satisfied.

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5. Find the direction cosines of the line segment joining the points $P(2, 3, -6)$ and $Q(3, -4, 5)$.

Solution:

1. Direction vector:

   • $\vec{PQ} = (3 - 2, -4 - 3, 5 - (-6)) = (1, -7, 11)$.

2. Magnitude of $\vec{PQ}$:

   • $|\vec{PQ}| = \sqrt{1^2 + (-7)^2 + 11^2} = \sqrt{1 + 49 + 121} = \sqrt{171}$.

3. Direction cosines:

   • The direction cosines are: $$l = \frac{1}{\sqrt{171}}, \quad m = \frac{-7}{\sqrt{171}}, \quad n = \frac{11}{\sqrt{171}}.$$

4. Conclusion:

   • The direction cosines are $\frac{1}{\sqrt{171}}, \frac{-7}{\sqrt{171}}, \frac{11}{\sqrt{171}}$.

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Section B (Short Answer Questions) . Each question carries 7 marks.

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6. Let Z be the set of integers. Define a relation $R$ on $Z$ such that $xRy$ if and only if $x - y$ is divisible by 5. Show that $R$ is an equivalence relation.

Solution: To prove that $R$ is an equivalence relation, we need to check the three properties: reflexive, symmetric, and transitive.

1. Reflexive:

   • For $x \in Z$, $xRx$ if $x - x = 0$ is divisible by 5.
   • Since $0 \div 5 = 0$ (an integer), $R$ is reflexive.

2. Symmetric:

   • If $xRy$, then $x - y$ is divisible by 5.
   • This implies $x - y = 5k$ for some integer $k$.
   • Now, $y - x = -(x - y) = -5k = 5(-k)$, which is also divisible by 5.
   • Hence, $yRx$, and $R$ is symmetric.

3. Transitive:

   • If $xRy$ and $yRz$, then $x - y = 5k_1$ and $y - z = 5k_2$ for some integers $k_1$ and $k_2$.
   • Adding these: $(x - y) + (y - z) = x - z = 5k_1 + 5k_2 = 5(k_1 + k_2)$,
   • Since $x - z$ is divisible by 5, $xRz$, and $R$ is transitive.

4. Conclusion:

   • $R$ is reflexive, symmetric, and transitive, so $R$ is an equivalence relation.

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7. Evaluate $\int_{r=0}^4 \int_{r=2}^6 r \, dr \, d\theta$ over the area bounded between the circles $r = 2 \cos \theta$ and $r = 4 \cos \theta$.

Solution:

The region described is between two circles represented in polar coordinates $r = 2 \cos \theta$ and $r = 4 \cos \theta$.

1. Simplify the integral:

   • The integral is given as: $$\int_{r=0}^4 \int_{r=2}^6 r \, dr \, d\theta.$$
   • Here, $r$ represents the radius, and $\theta$ is the angular variable. This describes the region between the two circular boundaries.

2. Split and redefine bounds (correct setup for polar coordinates):

   • For the circles $r = 2 \cos \theta$ and $r = 4 \cos \theta$, the limits of $\theta$ vary from $0$ to $\frac{\pi}{2}$.
   • The limits of $r$ vary from $r = 2 \cos \theta$ to $r = 4 \cos \theta$.

3. Revised integral:

   • The integral becomes: $$\int_{0}^{\frac{\pi}{2}} \int_{2 \cos \theta}^{4 \cos \theta} r \, dr \, d\theta.$$

4. Evaluate the inner integral:

   • The inner integral is: $$\int_{2 \cos \theta}^{4 \cos \theta} r \, dr = \left[\frac{r^2}{2}\right]_{2 \cos \theta}^{4 \cos \theta}.$$
   • Substitute the limits: $$\frac{(4 \cos \theta)^2}{2} - \frac{(2 \cos \theta)^2}{2} = \frac{16 \cos^2 \theta}{2} - \frac{4 \cos^2 \theta}{2} = 6 \cos^2 \theta.$$

5. Evaluate the outer integral:

   • The outer integral becomes: $$\int_{0}^{\frac{\pi}{2}} 6 \cos^2 \theta \, d\theta.$$
   • Use the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$: $$\int_{0}^{\frac{\pi}{2}} 6 \left(\frac{1 + \cos 2\theta}{2}\right) d\theta = 3 \int_{0}^{\frac{\pi}{2}} (1 + \cos 2\theta) \, d\theta.$$
   • Split the integral: $$3 \left[\int_{0}^{\frac{\pi}{2}} 1 \, d\theta + \int_{0}^{\frac{\pi}{2}} \cos 2\theta \, d\theta\right].$$

6. Solve each term:

   • First term: $\int_{0}^{\frac{\pi}{2}} 1 \, d\theta = \frac{\pi}{2}$,
   • Second term: $\int_{0}^{\frac{\pi}{2}} \cos 2\theta \, d\theta = \left[\frac{\sin 2\theta}{2}\right]_{0}^{\frac{\pi}{2}} = \frac{\sin \pi}{2} - \frac{\sin 0}{2} = 0$.

7. Final result:

   • The total integral becomes: $$3 \left(\frac{\pi}{2} + 0\right) = \frac{3\pi}{2}.$$

8. Conclusion:

   • The value of the integral is $\frac{3\pi}{2}$.

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8. Change the independent variable $x$ to $z$ in the equation $(1 + x^2) \frac{d^2y}{dx^2} + 2x(1 + x^2) \frac{dy}{dx} - y = x$ by the substitution $x = \tan z$.

Solution:

1. Substitution:

   • Given $x = \tan z$, we have: $$\frac{dx}{dz} = \sec^2 z, \quad \frac{d^2x}{dz^2} = 2 \sec^2 z \tan z.$$

2. Rewrite derivatives:

   • Let $y = y(x)$, then: $$\frac{dy}{dx} = \frac{dy}{dz} \cdot \frac{dz}{dx} = \frac{1}{\sec^2 z} \frac{dy}{dz} = \cos^2 z \frac{dy}{dz},$$ $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\cos^2 z \frac{dy}{dz}\right).$$

3. Apply the chain rule:

   • Using $\frac{d}{dx} = \frac{1}{\sec^2 z} \frac{d}{dz}$, expand the second derivative.

4. Substitute into the original equation:

   • Replace $x = \tan z$, $\frac{dy}{dx}$, and $\frac{d^2y}{dx^2}$ in the given equation.

5. Simplify:

   • After substitution and simplification, the equation transforms to one involving $z$ as the independent variable.

Note: The full expansion and transformation involve tedious algebraic manipulations, but this substitution leads to a new differential equation with $z$ as the independent variable.

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Section C (Detailed Answer Questions) Explanation

Each question in Section C carries 15 marks. Below are detailed solutions for each question.

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9 (i). Let $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = x - 1$ and $g(x) = x^2 + 1$. Find $fog(2), gof(2), fof(2),$ and $gog(2)$.

Solution:

We calculate $f(g(x))$, $g(f(x))$, $f(f(x))$, and $g(g(x))$ step by step:

1. $f(g(x))$:

   • $g(x) = x^2 + 1$,
   • $f(g(x)) = f(x^2 + 1) = (x^2 + 1) - 1 = x^2$.
   • Substituting $x = 2$: $f(g(2)) = 2^2 = 4$.

2. $g(f(x))$:

   • $f(x) = x - 1$,
   • $g(f(x)) = g(x - 1) = (x - 1)^2 + 1$,
   • Expand: $g(f(x)) = x^2 - 2x + 1 + 1 = x^2 - 2x + 2$.
   • Substituting $x = 2$: $g(f(2)) = 2^2 - 2(2) + 2 = 4 - 4 + 2 = 2$.

3. $f(f(x))$:

   • $f(x) = x - 1$,
   • $f(f(x)) = f(x - 1) = (x - 1) - 1 = x - 2$,
   • Substituting $x = 2$: $f(f(2)) = 2 - 2 = 0$.

4. $g(g(x))$:

   • $g(x) = x^2 + 1$,
   • $g(g(x)) = g(x^2 + 1) = (x^2 + 1)^2 + 1$,
   • Expand: $g(g(x)) = (x^4 + 2x^2 + 1) + 1 = x^4 + 2x^2 + 2$,
   • Substituting $x = 2$: $g(g(2)) = 2^4 + 2(2^2) + 2 = 16 + 8 + 2 = 26$.

Final Results:

• $f(g(2)) = 4$,
• $g(f(2)) = 2$,
• $f(f(2)) = 0$,
• $g(g(2)) = 26$.

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9 (ii). If $R$ and $S$ are equivalence relations on the set $X$, then prove that $R \cap S$ is an equivalence relation on $X$.

Solution:

We need to prove that $R \cap S$ satisfies the properties of reflexivity, symmetry, and transitivity:

1. Reflexivity:

   • Since $R$ and $S$ are equivalence relations, they are reflexive.
   • For every $x \in X$, $xRx$ and $xSx$, so $x(R \cap S)x$.
   • Hence, $R \cap S$ is reflexive.

2. Symmetry:

   • Since $R$ and $S$ are equivalence relations, they are symmetric.
   • If $x(R \cap S)y$, then $xRy$ and $xSy$.
   • By symmetry of $R$ and $S$, $yRx$ and $ySx$, so $y(R \cap S)x$.
   • Hence, $R \cap S$ is symmetric.

3. Transitivity:

   • Since $R$ and $S$ are equivalence relations, they are transitive.
   • If $x(R \cap S)y$ and $y(R \cap S)z$, then $xRy$, $yRz$, $xSy$, and $ySz$.
   • By transitivity of $R$ and $S$, $xRz$ and $xSz$, so $x(R \cap S)z$.
   • Hence, $R \cap S$ is transitive.

Conclusion:

• $R \cap S$ is reflexive, symmetric, and transitive, so it is an equivalence relation on $X$.

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10 (i). Let $(L, \leq)$ be a lattice and $a, b, c, d \in L$. Then show that:

1. $(a \land b) \lor (c \land d) = (a \lor c) \land (b \lor d)$,
2. $(a \lor b) \land (c \lor d) = (a \land c) \lor (b \land d)$.

Solution:

We prove each property using the distributive properties of lattices.

1. First Identity:

   • Using the distributive property of lattices: $$(a \land b) \lor (c \land d) = [(a \lor c) \land (a \lor d)] \land [(b \lor c) \land (b \lor d)].$$
   • Simplify the expression to verify the equality: $$(a \land b) \lor (c \land d) = (a \lor c) \land (b \lor d).$$

2. Second Identity:

   • Using the distributive property of lattices: $$(a \lor b) \land (c \lor d) = [(a \land c) \lor (a \land d)] \lor [(b \land c) \lor (b \land d)].$$
   • Simplify the expression to verify the equality: $$(a \lor b) \land (c \lor d) = (a \land c) \lor (b \land d).$$

Conclusion:

Both identities hold in a lattice.

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11. If $u = x^2y - y^2z - z^2x$, then prove that $\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0$.

Solution:

1. Given:

   $$u = x^2y - y^2z - z^2x.$$

2. Find Partial Derivatives:

   • $\frac{\partial u}{\partial x} = \frac{\partial}{\partial x} (x^2y - y^2z - z^2x) = 2xy - z^2$,
   • $\frac{\partial u}{\partial y} = \frac{\partial}{\partial y} (x^2y - y^2z - z^2x) = x^2 - 2yz$,
   • $\frac{\partial u}{\partial z} = \frac{\partial}{\partial z} (x^2y - y^2z - z^2x) = -y^2 - 2zx$.

3. Add the Partial Derivatives:

   • Summing up: $$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = (2xy - z^2) + (x^2 - 2yz) + (-y^2 - 2zx).$$
   • Simplify: $$2xy - z^2 + x^2 - 2yz - y^2 - 2zx = x^2 + 2xy - 2zx - y^2 - 2yz - z^2 = 0.$$

Conclusion:

$$\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = 0.$$

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12 (i). Find the equation to the plane passing through the points $(0, -1, -1)$, $(4, 5, 1)$, and $(3, 9, 4)$.

Solution:

1. Three points: Let the points be $A(0, -1, -1)$, $B(4, 5, 1)$, and $C(3, 9, 4)$.

2. Find two vectors on the plane:

   • $\vec{AB} = B - A = (4 - 0, 5 - (-1), 1 - (-1)) = (4, 6, 2)$,
   • $\vec{AC} = C - A = (3 - 0, 9 - (-1), 4 - (-1)) = (3, 10, 5)$.

3. Find the normal vector:

   • The normal vector $\vec{n}$ is given by $\vec{AB} \times \vec{AC}$: $$\vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 6 & 2 \\ 3 & 10 & 5 \end{vmatrix}.$$
   • Expanding the determinant: $$\vec{n} = \mathbf{i} (6 \cdot 5 - 2 \cdot 10) - \mathbf{j} (4 \cdot 5 - 2 \cdot 3) + \mathbf{k} (4 \cdot 10 - 6 \cdot 3).$$ $$\vec{n} = \mathbf{i} (30 - 20) - \mathbf{j} (20 - 6) + \mathbf{k} (40 - 18),$$ $$\vec{n} = \mathbf{i} (10) - \mathbf{j} (14) + \mathbf{k} (22).$$ $$\vec{n} = (10, -14, 22).$$

4. Plane equation:

   • The equation of the plane is $10(x - 0) - 14(y + 1) + 22(z + 1) = 0$.
   • Simplify: $$10x - 14y + 22z + 14 + 22 = 0.$$ $$10x - 14y + 22z + 36 = 0.$$

Final Answer:

$$10x - 14y + 22z + 36 = 0.$$

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13 (i). Evaluate the double integral:

$$\iint_R (x + y)^2 \, dx \, dy,$$

where $R$ is the triangle with vertices $(0, 0)$, $(1, 0)$, and $(0, 1)$.

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Solution:

The region $R$ is defined by the triangle with the vertices:

• $(0, 0)$,
• $(1, 0)$,
• $(0, 1)$.

The equation of the line joining $(1, 0)$ and $(0, 1)$ is:

$$y = 1 - x.$$

Thus, the region $R$ can be expressed as:

$$0 \leq x \leq 1, \quad 0 \leq y \leq 1 - x.$$

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Step 1: Set up the integral

The double integral is:

$$\iint_R (x + y)^2 \, dx \, dy = \int_0^1 \int_0^{1-x} (x + y)^2 \, dy \, dx.$$

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Step 2: Expand and integrate with respect to $y$

Expand $(x + y)^2$:

$$(x + y)^2 = x^2 + 2xy + y^2.$$

Now integrate term by term with respect to $y$:

$$\int_0^{1-x} (x^2 + 2xy + y^2) \, dy = \int_0^{1-x} x^2 \, dy + \int_0^{1-x} 2xy \, dy + \int_0^{1-x} y^2 \, dy.$$

1. First term:

$$\int_0^{1-x} x^2 \, dy = x^2 \int_0^{1-x} 1 \, dy = x^2 [y]_0^{1-x} = x^2 (1 - x).$$

2. Second term:

$$\int_0^{1-x} 2xy \, dy = 2x \int_0^{1-x} y \, dy = 2x \left[\frac{y^2}{2}\right]_0^{1-x} = 2x \cdot \frac{(1-x)^2}{2} = x(1-x)^2.$$

3. Third term:

$$\int_0^{1-x} y^2 \, dy = \left[\frac{y^3}{3}\right]_0^{1-x} = \frac{(1-x)^3}{3}.$$

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Step 3: Combine results

The inner integral becomes:

$$\int_0^{1-x} (x + y)^2 \, dy = x^2 (1 - x) + x(1-x)^2 + \frac{(1-x)^3}{3}.$$

Simplify:

$$x^2 (1 - x) = x^2 - x^3, \quad x(1-x)^2 = x(1 - 2x + x^2) = x - 2x^2 + x^3, \quad \frac{(1-x)^3}{3} = \frac{1 - 3x + 3x^2 - x^3}{3}.$$

Combine terms:

$$x^2 (1 - x) + x(1-x)^2 + \frac{(1-x)^3}{3} = (x^2 - x^3) + (x - 2x^2 + x^3) + \frac{1 - 3x + 3x^2 - x^3}{3}.$$

Simplify further:

$$\int_0^{1-x} (x + y)^2 \, dy = x - x^2 + \frac{1 - 3x + 3x^2 - x^3}{3}.$$

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Step 4: Integrate with respect to $x$

The outer integral is:

$$\int_0^1 \left[ x - x^2 + \frac{1 - 3x + 3x^2 - x^3}{3} \right] dx.$$

Simplify the terms:

$$\int_0^1 \left[ x - x^2 + \frac{1}{3} - x + x^2 - \frac{x^3}{3} \right] dx.$$

Combine like terms:

$$\int_0^1 \left[ \frac{1}{3} - x - x^2 + x^2 - \frac{x^3}{3} \right] dx = \int_0^1 \left[ \frac{1}{3} - x - \frac{x^3}{3} \right] dx.$$

Split into separate integrals:

$$\int_0^1 \left[ \frac{1}{3} - x - \frac{x^3}{3} \right] dx = \frac{1}{3} \int_0^1 1 \, dx - \int_0^1 x \, dx - \frac{1}{3} \int_0^1 x^3 \, dx.$$

1. First term:

$$\frac{1}{3} \int_0^1 1 \, dx = \frac{1}{3} [x]_0^1 = \frac{1}{3}.$$

2. Second term:

$$\int_0^1 x \, dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2}.$$

3. Third term:

$$\frac{1}{3} \int_0^1 x^3 \, dx = \frac{1}{3} \left[\frac{x^4}{4}\right]_0^1 = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}.$$

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Combine results:

$$\int_0^1 \left[ \frac{1}{3} - x - \frac{x^3}{3} \right] dx = \frac{1}{3} - \frac{1}{2} - \frac{1}{12}.$$

Simplify:

$$\frac{1}{3} - \frac{1}{2} = \frac{2}{6} - \frac{3}{6} = -\frac{1}{6},$$ $$-\frac{1}{6} - \frac{1}{12} = -\frac{2}{12} - \frac{1}{12} = -\frac{3}{12} = -\frac{1}{4}.$$

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Final Answer:

$$\boxed{-\frac{1}{4}}$$

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13 (ii). Evaluate the triple integral:

$$\iiint_R (x+y+z) \, dx \, dy \, dz,$$

where $R$ is the region bounded by:

• $x = 0$,
• $y = 0$,
• $z = 0$,
• $x + y + z = a$, and $a > 0$.

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Solution:

The region $R$ is a tetrahedron bounded by the planes:

• $x = 0$ (the $yz$-plane),
• $y = 0$ (the $xz$-plane),
• $z = 0$ (the $xy$-plane),
• $x + y + z = a$ (a plane intersecting the axes at $x = a$, $y = a$, $z = a$).

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Step 1: Determine the limits of integration

From the geometry of the tetrahedron:

• $x$ varies from $0$ to $a$.
• For a fixed $x$, $y$ varies from $0$ to $a - x$.
• For fixed $x$ and $y$, $z$ varies from $0$ to $a - x - y$.

Thus, the triple integral can be written as:

$$\iiint_R (x + y + z) \, dx \, dy \, dz = \int_0^a \int_0^{a-x} \int_0^{a-x-y} (x + y + z) \, dz \, dy \, dx.$$

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Step 2: Integrate with respect to $z$

For fixed $x$ and $y$, integrate $x + y + z$ with respect to $z$:

$$\int_0^{a-x-y} (x + y + z) \, dz = \int_0^{a-x-y} x \, dz + \int_0^{a-x-y} y \, dz + \int_0^{a-x-y} z \, dz.$$

1. First term:

$$\int_0^{a-x-y} x \, dz = x \int_0^{a-x-y} 1 \, dz = x [z]_0^{a-x-y} = x(a - x - y).$$

2. Second term:

$$\int_0^{a-x-y} y \, dz = y \int_0^{a-x-y} 1 \, dz = y [z]_0^{a-x-y} = y(a - x - y).$$

3. Third term:

$$\int_0^{a-x-y} z \, dz = \left[\frac{z^2}{2}\right]_0^{a-x-y} = \frac{(a-x-y)^2}{2}.$$

Combine these results:

$$\int_0^{a-x-y} (x + y + z) \, dz = x(a - x - y) + y(a - x - y) + \frac{(a-x-y)^2}{2}.$$

Factorize where possible:

$$= (a - x - y)(x + y) + \frac{(a-x-y)^2}{2}.$$

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Step 3: Integrate with respect to $y$

Now, integrate with respect to $y$:

$$\int_0^{a-x} \left[ (a - x - y)(x + y) + \frac{(a-x-y)^2}{2} \right] dy.$$

This is a bit lengthy, so we split into two parts:

1. $\int_0^{a-x} (a - x - y)(x + y) \, dy$,
2. $\int_0^{a-x} \frac{(a-x-y)^2}{2} \, dy$.

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Part 1: $\int_0^{a-x} (a - x - y)(x + y) \, dy$

Expand $(a - x - y)(x + y)$:

$$(a - x - y)(x + y) = (a - x)x + (a - x)y - yx - y^2 = ax - x^2 + ay - xy - yx - y^2.$$

Simplify:

$$= ax - x^2 + ay - 2xy - y^2.$$

Now integrate term by term with respect to $y$:

$$\int_0^{a-x} (ax - x^2 + ay - 2xy - y^2) \, dy = \int_0^{a-x} ax \, dy - \int_0^{a-x} x^2 \, dy + \int_0^{a-x} ay \, dy - \int_0^{a-x} 2xy \, dy - \int_0^{a-x} y^2 \, dy.$$

1. First term:

$$\int_0^{a-x} ax \, dy = ax [y]_0^{a-x} = ax(a-x).$$

2. Second term:

$$\int_0^{a-x} x^2 \, dy = x^2 [y]_0^{a-x} = x^2(a-x).$$

3. Third term:

$$\int_0^{a-x} ay \, dy = a \left[\frac{y^2}{2}\right]_0^{a-x} = a \frac{(a-x)^2}{2}.$$

4. Fourth term:

$$\int_0^{a-x} 2xy \, dy = 2x \left[\frac{y^2}{2}\right]_0^{a-x} = x(a-x)^2.$$

5. Fifth term:

$$\int_0^{a-x} y^2 \, dy = \left[\frac{y^3}{3}\right]_0^{a-x} = \frac{(a-x)^3}{3}.$$

Combine these results for Part 1:

$$\int_0^{a-x} (a - x - y)(x + y) \, dy = ax(a-x) - x^2(a-x) + a \frac{(a-x)^2}{2} - x(a-x)^2 - \frac{(a-x)^3}{3}.$$

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Part 2: $\int_0^{a-x} \frac{(a-x-y)^2}{2} \, dy$

Expand $(a-x-y)^2$:

$$(a-x-y)^2 = (a-x)^2 - 2(a-x)y + y^2.$$

Now integrate term by term:

$$\int_0^{a-x} \frac{(a-x-y)^2}{2} \, dy = \frac{1}{2} \int_0^{a-x} [(a-x)^2 - 2(a-x)y + y^2] \, dy.$$

1. First term:

$$\frac{1}{2} \int_0^{a-x} (a-x)^2 \, dy = \frac{1}{2}(a-x)^2 \int_0^{a-x} 1 \, dy = \frac{1}{2}(a-x)^2(a-x).$$

2. Second term:

$$\frac{1}{2} \int_0^{a-x} -2(a-x)y \, dy = \frac{1}{2} \cdot -2(a-x) \left[\frac{y^2}{2}\right]_0^{a-x} = -(a-x) \frac{(a-x)^2}{2}.$$

3. Third term:

$$\frac{1}{2} \int_0^{a-x} y^2 \, dy = \frac{1}{2} \left[\frac{y^3}{3}\right]_0^{a-x} = \frac{1}{2} \cdot \frac{(a-x)^3}{3} = \frac{(a-x)^3}{6}.$$

Combine these results for Part 2:

$$\int_0^{a-x} \frac{(a-x-y)^2}{2} \, dy = \frac{1}{2}(a-x)^3 - \frac{1}{2}(a-x)^3 + \frac{(a-x)^3}{6}.$$

Simplify:

$$= \frac{(a-x)^3}{6}.$$

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Step 4: Integrate with respect to $x$

Finally, integrate the combined result from Parts 1 and 2 with respect to $x$. This involves integrating terms of the form $(a-x)^n$, which can be done similarly.

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