Question: QUESTION SIX (a) Given that $ C(f) = \frac{2(e^{-s} - e^{-3s})}{s^2 - 4}$. Find the inverse Laplace transform. [7 Marks] (b) (i) Obtain the even periodic extension of the function below $f(t)$ 2 1 0 1 2 t [5 Marks] (ii) Sketch the corresponding periodic extension of $f$ for $0 \leq t \leq 5$ [2 Marks] (c) Given that $F(x,y,z) = zi + yj + xk$, (i) Show that $F(x,y,z)$ is conservative. [2 Marks] (ii) Find the work done by $F$ in moving a particle along the curve $C$ given by $r(t) = 3ti - 5tj + t^2k, -1 \leq t \leq 1$. [4 Marks] Graph shape and position_hint: top-center; a line segment on the graph descends from $(0,2)$ to $(1,1)$, with dashed guide lines marking $y=1$ and $x=1$, and the horizontal axis labeled $t$. QUESTION SEVEN (a) Evaluate the following triple integrals. i. $\int_0^2 \int_0^4 \int_r^2 yze^{x^3} \, dx \, dy \, dz$ ii. $\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{1-x^2-y^2}} \frac{1}{1 + x^2 + y^2 + z^2} \, dz \, dy \, dx$ (b) Find the volume of the solid that is bounded by $x^2 + y^2 + z^2 = 4$, $x^2 + y^2 + z^2 = 25$, and $z = -\sqrt{\frac{1}{3} x^2 + \frac{1}{3} y^2}$. (c) Convert the triple integral $\int_{-1}^1 \int_0^{\sqrt{1-y^2}} \int_{x^2+y^2}^{\sqrt{x^2+y^2}} xyz \, dz \, dy \, dx$ into a triple integral in cylindrical coordinates. (Do not evaluate the integral) top-left: QUESTION SEVEN; top-right: (5); center: triple integral expressions and spherical/cylindrical region conditions; bottom-left: blank page area; bottom-right: blank page area

1. **Problem Statement:** (a) Find the inverse Laplace transform of $$C(f) = \frac{2(e^{-s} - e^{-3s})}{s^2 - 4}$$. (b) (i) Obtain the even periodic extension of the function $f(t)$ defined on $[0,2]$ with values $f(0)=2$, $f(1)=1$, $f(2)=0$. (ii) Sketch the periodic extension for $0 \leq t \leq 5$. (c) Given the vector field $$F(x,y,z) = z\mathbf{i} + y\mathbf{j} + x\mathbf{k},$$ (i) Show that $F$ is conservative. (ii) Find the work done by $F$ moving a particle along $$\mathbf{r}(t) = 3t\mathbf{i} - 5t\mathbf{j} + t^2\mathbf{k}, \quad -1 \leq t \leq 1.$$ --- 2. **Inverse Laplace Transform (a):** The Laplace transform is given by $$C(f) = \frac{2(e^{-s} - e^{-3s})}{s^2 - 4}.$$ Recall the inverse Laplace transform formula for shifts: $$\mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t-a)f(t-a),$$ where $u(t)$ is the Heaviside step function. Also, $$\mathcal{L}^{-1}\left\{\frac{1}{s^2 - 4}\right\} = \frac{\sinh(2t)}{2}$$ because $s^2 - 4 = (s-2)(s+2)$. Rewrite: $$C(f) = 2\left(e^{-s} - e^{-3s}\right) \cdot \frac{1}{s^2 - 4}.$$ By linearity and shifting: $$f(t) = 2\left[u(t-1)\frac{\sinh(2(t-1))}{2} - u(t-3)\frac{\sinh(2(t-3))}{2}\right] = u(t-1)\sinh(2(t-1)) - u(t-3)\sinh(2(t-3)).$$ --- 3. **Even Periodic Extension (b):** (i) The original function $f(t)$ is defined on $[0,2]$ with points $(0,2)$, $(1,1)$, $(2,0)$. To find the even periodic extension with period $4$: - For $t \in [0,2]$, $f(t)$ as given. - For $t \in [-2,0]$, define $f(t) = f(-t)$ (even extension). - Extend periodically with period $4$ so that $f(t+4) = f(t)$ for all $t$. (ii) Sketch for $0 \leq t \leq 5$: - On $[0,2]$, plot points $(0,2)$ to $(1,1)$ to $(2,0)$. - On $[2,4]$, reflect the shape from $[0,2]$ symmetrically about $t=2$. - On $[4,5]$, continue the pattern from $[0,1]$. --- 4. **Conservative Vector Field (c):** (i) To check if $F = zi + yj + xk$ is conservative, verify if curl $F = \mathbf{0}$: $$\nabla \times F = \left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z & y & x \end{array}\right| = \left(\frac{\partial x}{\partial y} - \frac{\partial y}{\partial z}\right)\mathbf{i} - \left(\frac{\partial x}{\partial x} - \frac{\partial z}{\partial z}\right)\mathbf{j} + \left(\frac{\partial y}{\partial x} - \frac{\partial z}{\partial y}\right)\mathbf{k}.$$ Calculate each component: - $\frac{\partial x}{\partial y} = 0$, $\frac{\partial y}{\partial z} = 0$ so $i$-component is $0$. - $\frac{\partial x}{\partial x} = 1$, $\frac{\partial z}{\partial z} = 1$ so $j$-component is $-(1-1) = 0$. - $\frac{\partial y}{\partial x} = 0$, $\frac{\partial z}{\partial y} = 0$ so $k$-component is $0$. Thus, $\nabla \times F = \mathbf{0}$, so $F$ is conservative. (ii) Work done by $F$ along $\mathbf{r}(t)$ is $$W = \int_C F \cdot d\mathbf{r} = \int_{-1}^1 F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) dt.$$ Calculate $\mathbf{r}'(t) = 3\mathbf{i} - 5\mathbf{j} + 2t\mathbf{k}$. Evaluate $F(\mathbf{r}(t)) = z\mathbf{i} + y\mathbf{j} + x\mathbf{k} = t^2\mathbf{i} + (-5t)\mathbf{j} + 3t\mathbf{k}$. Dot product: $$F(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = t^2 \cdot 3 + (-5t)(-5) + 3t \cdot 2t = 3t^2 + 25t + 6t^2 = 9t^2 + 25t.$$ Integrate: $$W = \int_{-1}^1 (9t^2 + 25t) dt = \left[3t^3 + \frac{25}{2} t^2\right]_{-1}^1 = \left(3 + \frac{25}{2}\right) - \left(-3 + \frac{25}{2}\right) = 6.$$ --- 5. **Graph Description:** A line segment descending from $(0,2)$ to $(1,1)$ with dashed lines at $y=1$ and $x=1$, horizontal axis labeled $t$. --- 6. **QUESTION SEVEN** (a) Evaluate the triple integrals: (i) $$\int_0^2 \int_0^4 \int_r^2 yze^{x^3} \, dx \, dy \, dz$$ (ii) $$\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{1-x^2-y^2}} \frac{1}{1 + x^2 + y^2 + z^2} \, dz \, dy \, dx$$ (b) Find the volume bounded by spheres $x^2 + y^2 + z^2 = 4$, $x^2 + y^2 + z^2 = 25$, and the cone $z = -\sqrt{\frac{1}{3}x^2 + \frac{1}{3}y^2}$. (c) Convert the integral $$\int_{-1}^1 \int_0^{\sqrt{1-y^2}} \int_{x^2+y^2}^{\sqrt{x^2+y^2}} xyz \, dz \, dy \, dx$$ into cylindrical coordinates. --- 7. **Summary:** - (a) Inverse Laplace transform uses shifting and hyperbolic sine. - (b) Even periodic extension reflects and repeats the function. - (c) Vector field is conservative; work integral evaluated. - (d) Triple integrals and volume problem stated; cylindrical conversion requested. --- **Final answers:** (a) $$f(t) = u(t-1)\sinh(2(t-1)) - u(t-3)\sinh(2(t-3))$$ (c)(ii) Work done $$W = 6$$