AP Calculus AB Exam Question:

Consider the curve defined by the function $y = f(x)$, where $f(x)$ is a continuous function on the interval $[0, 2]$. The region bounded by the x-axis and this curve is revolved about the y-axis, creating a solid of revolution.

(a) Find the volume of the solid generated when the shaded region is revolved about the y-axis.

(b) The function $f(x)$ is defined as follows:

$$f(x) = \begin{cases} x^2 & 0 \leq x < 1 \\ 2-x & 1 \leq x \leq 2 \\ \end{cases}$$

Calculate the volume of the solid of revolution generated when the shaded region is revolved about the y-axis.

(Note: The formula for the volume of a solid of revolution, when revolved about the y-axis, is given by $V = \pi\int_a^b f(x)^2 dx$, where $a$ and $b$ are the x-values determining the region of revolution.)

Answer:

(a) To find the volume of the solid generated by revolving the shaded region about the y-axis, we'll make use of the formula $V = \pi\int_a^b f(x)^2 dx$, where $a$ and $b$ are the x-values that determine the region of revolution.

In this case, the region of revolution is bounded by the x-axis and the curve $y = f(x)$. The graph of $y = f(x)$ intersects the x-axis at $x = 0$ and $x = 2$, so $a = 0$ and $b = 2$.

We need to find the function $f(x)$ in terms of $y$ to rewrite the integral:

$$x = \begin{cases} \sqrt{y} & 0 \leq y < 1 \\ 2-y & 1 \leq y \leq 4 \\ \end{cases}$$

Now, let's rewrite the integral in terms of $y$:

$$V = \pi\int_0^4 x^2 dy$$

Using the transformation $x = \sqrt{y}$ for $0 \leq y < 1$:

$$V = \pi\int_0^1 (\sqrt{y})^2 dy = \pi\int_0^1 y dy$$

Using the transformation $x = 2-y$ for $1 \leq y \leq 4$:

$$V = \pi\int_1^4 (2-y)^2 dy = \pi\int_1^4 (y^2 - 4y + 4) dy$$

Simplifying the integrals:

$$V = \pi\left[\frac{y^2}{2}\right]_0^1 + \pi\left[\frac{y^3}{3} - 2y^2 + 4y\right]_1^4$$

$$V = \pi\left(\frac{1}{2}\right) + \pi\left(\frac{64}{3} - 32 + 16 - \left(\frac{1}{3} - 2 + 4\right)\right)$$

$$V = \pi\left(\frac{1}{2}\right) + \pi\left(\frac{64}{3} - 32 + 16 - \frac{5}{3} + 2 - 4\right)$$

$$V = \pi\left(\frac{1}{2}\right) + \pi\left(\frac{27}{3}\right)$$

$$V = \pi\left(\frac{1}{2} + 9\right) = \pi\left(\frac{19}{2}\right)$$

Therefore, the volume of the solid generated by revolving the shaded region about the y-axis is $\pi\left(\frac{19}{2}\right)$.

(b) Using the function $f(x)$ provided, we can use the same formula $V = \pi\int_a^b f(x)^2 dx$ to find the volume of the solid generated when the shaded region is revolved about the y-axis.

Considering how $f(x)$ is defined, we have:

$$V = \pi\int_0^1 (x^2)^2 dx + \pi\int_1^2 (2-x)^2 dx$$

Simplifying the integral expression:

$$V = \pi\int_0^1 x^4 dx + \pi\int_1^2 (x^2 - 4x + 4) dx$$

$$V = \pi\left[\frac{x^5}{5}\right]_0^1 + \pi\left[\frac{x^3}{3} - 2x^2 + 4x\right]_1^2$$

$$V = \pi\left(\frac{1}{5}\right) + \pi\left(\frac{8}{3} - 8 + 8 - \frac{1}{3} + 2 - 4\right)$$

$$V = \pi\left(\frac{1}{5}\right) + \pi\left(\frac{16}{3} - 1 - 2\right)$$

$$V = \pi\left(\frac{1}{5}\right) + \pi\left(\frac{16}{3} - \frac{5}{3}\right)$$

$$V = \pi\left(\frac{1}{5}\right) + \pi\left(\frac{11}{3}\right)$$

$$V = \pi\left(\frac{3}{15}\right) + \pi\left(\frac{55}{15}\right)$$

$$V = \pi\left(\frac{3 + 55}{15}\right) = \pi\left(\frac{58}{15}\right)$$

Therefore, the volume of the solid of revolution generated when the shaded region is revolved about the y-axis is $\pi\left(\frac{58}{15}\right)$.