AP Calculus AB Practice Quiz: Volume with Disc Method: Revolving Around the $x$- or $y$-Axis

Correct Answer: A

When revolving a region around the $x$-axis using the disc method, the volume is given by the integral $V = \\pi \int_{a}^{b} [R(x)]^2 \,dx$. For the region bounded by $y = \\sqrt{x}$, the radius of each disc is $R(x) = y = \\sqrt{x}$. The region is bounded by $x=0$ (implied by the $x$-axis and the function's domain) and $x=4$. Therefore, the integral for the volume is $V = \\pi \int_{0}^{4} (\\sqrt{x})^2 \,dx = \\pi \int_{0}^{4} x \,dx$.

Correct Answer: D

To find the volume of a solid revolved around the $y$-axis using the disc method, the formula is $V = \\pi \int_{c}^{d} [R(y)]^2 \,dy$. The radius $R(y)$ is the horizontal distance from the $y$-axis to the curve, which is the $x$-value. We must express $x$ as a function of $y$. From $y = x^2$, we get $x = \\sqrt{y}$ (since we are in the first quadrant). The bounds of integration are along the $y$-axis, from $y=0$ to $y=9$. Therefore, the integral is $V = \\pi \int_{0}^{9} (\\sqrt{y})^2 \,dy = \\pi \int_{0}^{9} y \,dy$.

Correct Answer: D

Using the disc method for revolution about the $x$-axis, the volume is $V = \\pi \int_{a}^{b} [R(x)]^2 \,dx$. Here, the radius is $R(x) = 2x$, and the bounds are from $a=0$ to $b=3$. The integral is $V = \\pi \int_{0}^{3} (2x)^2 \,dx = \\pi \int_{0}^{3} 4x^2 \,dx$. Evaluating the integral gives $V = \\pi [\\frac{4x^3}{3}]_{0}^{3} = \\pi (\\frac{4(3)^3}{3} - 0) = \\pi (4 \\cdot 9) = 36\\pi$.

Correct Answer: C

The volume is found using the disc method: $V = \\pi \int_{a}^{b} [f(x)]^2 \,dx$. Here, $f(x) = \sec(x)$ and the interval is $[-\\frac{\\pi}{4}, \\frac{\\pi}{4}]$. The setup is $V = \\pi \int_{-\\pi/4}^{\\pi/4} (\sec(x))^2 \,dx$. The antiderivative of $\sec^2(x)$ is $\\tan(x)$. Evaluating the definite integral: $V = \\pi [\\tan(x)]_{-\\pi/4}^{\\pi/4} = \\pi (\\tan(\\frac{\\pi}{4}) - \\tan(-\\frac{\\pi}{4})) = \\pi (1 - (-1)) = \\pi(2) = 2\\pi$.

Correct Answer: C

For revolution about the $y$-axis, we use $V = \\pi \int_{c}^{d} [R(y)]^2 \,dy$. We need to express the radius $x$ as a function of $y$: $y = x^3 \implies x = y^{1/3}$. The bounds on the $y$-axis are from $c=0$ to $d=8$. The integral is $V = \\pi \int_{0}^{8} (y^{1/3})^2 \,dy = \\pi \int_{0}^{8} y^{2/3} \,dy$. Evaluating this gives: $V = \\pi [\\frac{3}{5}y^{5/3}]_{0}^{8} = \\frac{3\\pi}{5} (8^{5/3} - 0^{5/3}) = \\frac{3\\pi}{5} ((8^{1/3})^5) = \\frac{3\\pi}{5} (2^5) = \\frac{3\\pi}{5} (32) = \\frac{96\\pi}{5}$.

Correct Answer: A

The volume of a solid of revolution about the $x$-axis is given by the disc method formula $V = \\pi \int_{a}^{b} [R(x)]^2 \,dx$. In this case, the region is bounded by the function $y = \\frac{r}{h}x$, so the radius of a disc at any $x$ is $R(x) = \\frac{r}{h}x$. The integration is performed along the $x$-axis from $a=0$ to $b=h$. Plugging these into the formula gives $V = \\pi \int_{0}^{h} (\\frac{r}{h}x)^2 \,dx$. Option B is missing the square on the radius. Option C incorrectly sets up the integral for revolution around the $y$-axis. Option D is missing the factor of $\\pi$.

Correct Answer: A

The volume is given by $V = \\pi \int_{1}^{k} (\\frac{1}{\\sqrt{x}})^2 \,dx = \\pi \int_{1}^{k} \\frac{1}{x} \,dx$. The integral of $\\frac{1}{x}$ is $\\ln|x|$. So, $V = \\pi [\\ln|x|]_{1}^{k} = \\pi (\\ln(k) - \\ln(1)) = \\pi \\ln(k)$, since $k>1$ and $\\ln(1)=0$. We are given that the volume is $3\\pi$. Setting them equal: $\\pi \\ln(k) = 3\\pi$. Dividing by $\\pi$ gives $\\ln(k) = 3$. To solve for $k$, we exponentiate both sides: $k = e^3$.