Question: 3.1 Consider the curves $y = \cos x$ and $y = \frac{2x}{\pi} - 1$, on the interval $0 \leq x \leq \pi$. The region enclosed by these curves will be investigated using definite integration. Adapted from Stewart, J., Clegg, D.K and Watson, S. 2021. Calculus: Early Transcendentals. 9th Metric Edition. Cengage Learning. Unless otherwise stated, give all answers in exact form and show all necessary mathematical working. a) Determine the point of intersection of the two curves in the given interval. Verify your answer by substitution into both equations. (3 Marks) b) Using the point of intersection obtained in 3.1a), determine which function forms the upper boundary on each resulting subinterval. Justify your answer mathematically. (3 Marks) c) Hence, determine the exact total area enclosed between the two curves on $0 \leq x \leq \pi$. (4 Marks) 3.2 Consider the region bounded by $y = x(6 - x)$, $y = 0$, $x = 1$, $x = 5$. The region is rotated through one complete revolution about the vertical line $x = -2$. a) Describe the three-dimensional solid generated when this region is revolved through one complete revolution about the vertical line $x = -2$. (2 Marks) b) Select an appropriate integration method to determine the volume of the solid generated. Briefly justify your choice. Set up and evaluate the appropriate definite integral to obtain the exact volume. Show all necessary mathematical working. (6 Marks) c) Suppose the same region is instead rotated about the y-axis. Without evaluating another integral, explain how the relevant dimension of the solid element would change and write the corresponding volume integral. (2 Marks) 3.3 Consider the region bounded by $y = \sqrt{x + 1}$, $y = 0$, $x = 0$, $x = 3$. The region is rotated through one complete revolution about the x-axis. a) For a representative vertical strip at $x$, determine the relevant dimension(s) of the solid element formed after rotation. (2 Marks) b) Select an appropriate integration method to determine the exact volume of the solid generated. Briefly justify your choice, then set up and evaluate the required definite integral. Show all necessary mathematical working. (5 Marks) c) If the same region were instead rotated about the line $y = -1$, identify the appropriate integration method and justify your choice.

### Problem 3.1 1. **State the problem:** Find the point(s) of intersection of the curves $y = \cos x$ and $y = \frac{2x}{\pi} - 1$ on $0 \leq x \leq \pi$, determine which curve is upper on subintervals, and find the total enclosed area. 2. **Find the point of intersection:** Set the two functions equal: $$\cos x = \frac{2x}{\pi} - 1$$ Rearranged: $$\cos x + 1 = \frac{2x}{\pi}$$ Multiply both sides by $\pi$: $$\pi (\cos x + 1) = 2x$$ This transcendental equation is solved exactly by inspection or numerically. Check $x = \frac{\pi}{2}$: $$\cos \frac{\pi}{2} = 0, \quad \frac{2 \cdot \frac{\pi}{2}}{\pi} - 1 = 1 - 1 = 0$$ Both sides equal 0, so the point of intersection is at: $$\boxed{\left(\frac{\pi}{2}, 0\right)}$$ 3. **Verify by substitution:** - For $y = \cos x$ at $x = \frac{\pi}{2}$: $$y = \cos \frac{\pi}{2} = 0$$ - For $y = \frac{2x}{\pi} - 1$ at $x = \frac{\pi}{2}$: $$y = \frac{2 \cdot \frac{\pi}{2}}{\pi} - 1 = 1 - 1 = 0$$ Both equal, confirming the intersection. 4. **Determine upper boundary on subintervals:** - On $[0, \frac{\pi}{2})$, test $x=0$: $$\cos 0 = 1, \quad \frac{2 \cdot 0}{\pi} - 1 = -1$$ Since $1 > -1$, $y=\cos x$ is upper. - On $(\frac{\pi}{2}, \pi]$, test $x=\pi$: $$\cos \pi = -1, \quad \frac{2 \pi}{\pi} - 1 = 2 - 1 = 1$$ Since $1 > -1$, $y=\frac{2x}{\pi} - 1$ is upper. 5. **Calculate the enclosed area:** The total area is the sum of integrals of the difference of upper minus lower function on each subinterval: $$\text{Area} = \int_0^{\frac{\pi}{2}} (\cos x - (\frac{2x}{\pi} - 1)) \, dx + \int_{\frac{\pi}{2}}^{\pi} \left(\frac{2x}{\pi} - 1 - \cos x\right) \, dx$$ 6. **Simplify integrands:** - First integral: $$\cos x - \frac{2x}{\pi} + 1$$ - Second integral: $$\frac{2x}{\pi} - 1 - \cos x$$ 7. **Evaluate first integral:** $$\int_0^{\frac{\pi}{2}} \cos x \, dx - \int_0^{\frac{\pi}{2}} \frac{2x}{\pi} \, dx + \int_0^{\frac{\pi}{2}} 1 \, dx$$ Calculate each: - $\int_0^{\frac{\pi}{2}} \cos x \, dx = \sin x \big|_0^{\frac{\pi}{2}} = 1 - 0 = 1$ - $\int_0^{\frac{\pi}{2}} \frac{2x}{\pi} \, dx = \frac{2}{\pi} \cdot \frac{x^2}{2} \big|_0^{\frac{\pi}{2}} = \frac{1}{\pi} \cdot \left(\frac{\pi^2}{4}\right) = \frac{\pi}{4}$ - $\int_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}$ Sum: $$1 - \frac{\pi}{4} + \frac{\pi}{2} = 1 + \frac{\pi}{4}$$ 8. **Evaluate second integral:** $$\int_{\frac{\pi}{2}}^{\pi} \frac{2x}{\pi} \, dx - \int_{\frac{\pi}{2}}^{\pi} 1 \, dx - \int_{\frac{\pi}{2}}^{\pi} \cos x \, dx$$ Calculate each: - $\int_{\frac{\pi}{2}}^{\pi} \frac{2x}{\pi} \, dx = \frac{2}{\pi} \cdot \frac{x^2}{2} \big|_{\frac{\pi}{2}}^{\pi} = \frac{1}{\pi} (\pi^2 - \frac{\pi^2}{4}) = \frac{1}{\pi} \cdot \frac{3\pi^2}{4} = \frac{3\pi}{4}$ - $\int_{\frac{\pi}{2}}^{\pi} 1 \, dx = \pi - \frac{\pi}{2} = \frac{\pi}{2}$ - $\int_{\frac{\pi}{2}}^{\pi} \cos x \, dx = \sin x \big|_{\frac{\pi}{2}}^{\pi} = 0 - 1 = -1$ Sum: $$\frac{3\pi}{4} - \frac{\pi}{2} - (-1) = \frac{3\pi}{4} - \frac{\pi}{2} + 1 = \frac{3\pi}{4} - \frac{2\pi}{4} + 1 = \frac{\pi}{4} + 1$$ 9. **Total enclosed area:** $$\text{Area} = \left(1 + \frac{\pi}{4}\right) + \left(\frac{\pi}{4} + 1\right) = 2 + \frac{\pi}{2}$$ ### Problem 3.2 1. **Describe the solid:** The region bounded by $y = x(6 - x)$, $y=0$, $x=1$, and $x=5$ is revolved about the vertical line $x = -2$. This generates a solid of revolution with circular cross-sections perpendicular to the axis of rotation. 2. **Integration method:** Use the **shell method** because the axis of rotation is vertical and the region is bounded in $x$. 3. **Set up the volume integral:** - Radius of shell at $x$ is distance from $x$ to axis $x = -2$: $$r = x - (-2) = x + 2$$ - Height of shell is $y = x(6 - x) = 6x - x^2$ - Volume: $$V = 2\pi \int_1^5 (\text{radius})(\text{height}) \, dx = 2\pi \int_1^5 (x + 2)(6x - x^2) \, dx$$ 4. **Expand integrand:** $$(x + 2)(6x - x^2) = 6x^2 - x^3 + 12x - 2x^2 = (6x^2 - 2x^2) + 12x - x^3 = 4x^2 + 12x - x^3$$ 5. **Evaluate integral:** $$V = 2\pi \int_1^5 (4x^2 + 12x - x^3) \, dx = 2\pi \left[ \frac{4x^3}{3} + 6x^2 - \frac{x^4}{4} \right]_1^5$$ Calculate each term at bounds: - At $x=5$: $$\frac{4 \cdot 125}{3} + 6 \cdot 25 - \frac{625}{4} = \frac{500}{3} + 150 - 156.25$$ - At $x=1$: $$\frac{4}{3} + 6 - \frac{1}{4} = \frac{4}{3} + 6 - 0.25$$ 6. **Compute difference:** $$\left(\frac{500}{3} + 150 - 156.25\right) - \left(\frac{4}{3} + 6 - 0.25\right) = \left(\frac{500}{3} - \frac{4}{3}\right) + (150 - 6) + (-156.25 + 0.25) = \frac{496}{3} + 144 - 156$$ Simplify: $$\frac{496}{3} + (144 - 156) = \frac{496}{3} - 12 = \frac{496}{3} - \frac{36}{3} = \frac{460}{3}$$ 7. **Final volume:** $$V = 2\pi \cdot \frac{460}{3} = \frac{920\pi}{3}$$ ### Problem 3.2c 1. **Rotation about y-axis:** The radius of the shell changes from $r = x + 2$ to $r = x$ (distance from $x$ to $x=0$). 2. **Volume integral:** $$V = 2\pi \int_1^5 x \cdot (6x - x^2) \, dx = 2\pi \int_1^5 (6x^2 - x^3) \, dx$$ ### Problem 3.3a 1. **Solid element dimensions:** For vertical strip at $x$, height is $y = \sqrt{x+1}$. 2. When rotated about x-axis, the radius of the disk is $r = y = \sqrt{x+1}$. ### Problem 3.3b 1. **Integration method:** Use the **disk method** because the solid is generated by rotating around the x-axis and the cross-sections perpendicular to x-axis are disks. 2. **Volume integral:** $$V = \pi \int_0^3 (\sqrt{x+1})^2 \, dx = \pi \int_0^3 (x+1) \, dx$$ 3. **Evaluate integral:** $$V = \pi \left[ \frac{x^2}{2} + x \right]_0^3 = \pi \left( \frac{9}{2} + 3 \right) = \pi \cdot \frac{15}{2} = \frac{15\pi}{2}$$ ### Problem 3.3c 1. **Rotation about $y = -1$:** The radius of disks changes to $r = y + 1 = \sqrt{x+1} + 1$. 2. **Integration method:** Still use the **disk method** because rotation is about a horizontal line parallel to x-axis. 3. **Justification:** The cross-section disks have radius equal to vertical distance from curve to line $y = -1$. --- **Final answers:** - 3.1a) Intersection at $\left(\frac{\pi}{2}, 0\right)$ - 3.1b) $y=\cos x$ upper on $[0, \frac{\pi}{2})$, $y=\frac{2x}{\pi} - 1$ upper on $(\frac{\pi}{2}, \pi]$ - 3.1c) Total enclosed area $= 2 + \frac{\pi}{2}$ - 3.2a) Solid of revolution generated by revolving region about vertical line $x=-2$ - 3.2b) Volume $= \frac{920\pi}{3}$ using shell method - 3.2c) Volume integral about y-axis: $V = 2\pi \int_1^5 x(6x - x^2) \, dx$ - 3.3a) Radius of disk $= \sqrt{x+1}$ - 3.3b) Volume $= \frac{15\pi}{2}$ using disk method - 3.3c) Rotation about $y=-1$ uses disk method with radius $r = \sqrt{x+1} + 1$