1. **Problem statement:** Find the value of $t$ where line $WY$ is normal to the curve $y=\frac{1}{2}x^{2}+1$ at point $B(2,4)$, then find the area of the shaded region bounded by the curve, line $BQ$ (vertical through $B$), and line $WY$, and finally find the volume generated when this shaded region is revolved about the $y$-axis. 2. **Step a: Find $t$ (the x-intercept of line $WY$ normal to the curve at $B$).** - The curve is $y=\frac{1}{2}x^{2}+1$. - The derivative (slope of tangent) is $y' = x$. - At $B(2,4)$, slope of tangent is $m_{tangent} = 2$. - Slope of normal line is $m_{normal} = -\frac{1}{m_{tangent}} = -\frac{1}{2}$. - Equation of normal line $WY$ passing through $B(2,4)$: $$y - 4 = -\frac{1}{2}(x - 2)$$ - Find $t$ where $WY$ meets the x-axis ($y=0$): $$0 - 4 = -\frac{1}{2}(t - 2) \implies -4 = -\frac{1}{2}t + 1 \implies -5 = -\frac{1}{2}t \implies t = 10$$ 3. **Step b: Find the area of the shaded region.** - The shaded region is bounded by the curve from $x=0$ to $x=2$, the vertical line $BQ$ at $x=2$, and the normal line $WY$ from $B$ to $Y(t,0)$. - The area under the curve from $x=0$ to $x=2$: $$A_1 = \int_0^2 \left(\frac{1}{2}x^{2} + 1\right) dx = \left[\frac{1}{6}x^{3} + x\right]_0^2 = \frac{8}{6} + 2 = \frac{4}{3} + 2 = \frac{10}{3}$$ - The area under the normal line $WY$ from $x=2$ to $x=10$: Equation of $WY$: $y = -\frac{1}{2}x + 5$ $$A_2 = \int_2^{10} \left(-\frac{1}{2}x + 5\right) dx = \left[-\frac{1}{4}x^{2} + 5x\right]_2^{10} = \left(-\frac{1}{4} \times 100 + 50\right) - \left(-\frac{1}{4} \times 4 + 10\right) = ( -25 + 50 ) - ( -1 + 10 ) = 25 - 9 = 16$$ - The total area of the shaded region is the sum of $A_1$ and $A_2$: $$A = \frac{10}{3} + 16 = \frac{10}{3} + \frac{48}{3} = \frac{58}{3}$$ 4. **Step c: Find the volume of the solid formed by revolving the shaded region about the y-axis.** - The volume is generated by revolving the region bounded by the curve and the normal line about the y-axis. - Use the shell method: volume $V = 2\pi \int_{x=0}^{x=10} x \cdot f(x) \, dx$ where $f(x)$ is the height of the region at $x$. - For $0 \leq x \leq 2$, height is the curve $y=\frac{1}{2}x^{2}+1$. - For $2 < x \leq 10$, height is the normal line $y = -\frac{1}{2}x + 5$. - Compute volume: $$V = 2\pi \left( \int_0^2 x \left(\frac{1}{2}x^{2} + 1\right) dx + \int_2^{10} x \left(-\frac{1}{2}x + 5\right) dx \right)$$ - Calculate first integral: $$\int_0^2 x \left(\frac{1}{2}x^{2} + 1\right) dx = \int_0^2 \left(\frac{1}{2}x^{3} + x\right) dx = \left[\frac{1}{8}x^{4} + \frac{1}{2}x^{2}\right]_0^2 = \left(\frac{1}{8} \times 16 + \frac{1}{2} \times 4\right) = 2 + 2 = 4$$ - Calculate second integral: $$\int_2^{10} x \left(-\frac{1}{2}x + 5\right) dx = \int_2^{10} \left(-\frac{1}{2}x^{2} + 5x\right) dx = \left[-\frac{1}{6}x^{3} + \frac{5}{2}x^{2}\right]_2^{10}$$ Evaluate: $$= \left(-\frac{1}{6} \times 1000 + \frac{5}{2} \times 100\right) - \left(-\frac{1}{6} \times 8 + \frac{5}{2} \times 4\right) = (-166.6667 + 250) - (-1.3333 + 10) = 83.3333 - 8.6667 = 74.6666$$ - Total volume: $$V = 2\pi (4 + 74.6666) = 2\pi \times 78.6666 = 157.3333\pi = \frac{472}{3}\pi$$ **Final answers:** - a. $t = 10$ - b. Area of shaded region $= \frac{58}{3}$ - c. Volume of solid $= \frac{472}{3}\pi$