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

Correct Answer: A

The region is bounded by the two curves which intersect at $x=0$ and $x=1$. When revolved around the $x$-axis, the cross-sections are ring-shaped (washers). The outer radius $R(x)$ is the distance from the $x$-axis to the upper curve, $y=x$. The inner radius $r(x)$ is the distance from the $x$-axis to the lower curve, $y=x^2$. The volume is found using the washer method formula $V = \\pi \int_{a}^{b} (R(x)^2 - r(x)^2) dx$. Therefore, the integral is $V = \\pi \int_{0}^{1} (x^2 - (x^2)^2) dx = \\pi \int_{0}^{1} (x^2 - x^4) dx$.

Correct Answer: C

To revolve around the $y$-axis, we must integrate with respect to $y$. The region spans from $y=0$ to $y=4$ (since $y=x^2$ and $x=2$). The cross-sections are washers. The outer radius $R(y)$ is the distance from the $y$-axis to the line $x=2$, so $R(y)=2$. The inner radius $r(y)$ is the distance from the $y$-axis to the curve $y=x^2$, which must be expressed as $x=\\sqrt{y}$, so $r(y)=\\sqrt{y}$. The volume is $V = \\pi \int_{0}^{4} (R(y)^2 - r(y)^2) dy = \\pi \int_{0}^{4} (2^2 - (\\sqrt{y})^2) dy = \\pi \int_{0}^{4} (4 - y) dy$.

Correct Answer: C

The intersection points of $y=x^2$ and $y=4$ are at $x=-2$ and $x=2$. When this region is revolved about the $x$-axis, the cross-sections are washers. The outer radius $R(x)$ is the distance from the $x$-axis to the farther curve, $y=4$. The inner radius $r(x)$ is the distance from the $x$-axis to the closer curve, $y=x^2$. Using the washer method, the volume is $V = \\pi \int_{-2}^{2} (R(x)^2 - r(x)^2) dx = \\pi \int_{-2}^{2} (4^2 - (x^2)^2) dx = \\pi \int_{-2}^{2} (16 - x^4) dx$.

Correct Answer: C

Using the washer method for revolution about the $y$-axis, the volume integral is $V = \\pi \int_{0}^{4} (R(y)^2 - r(y)^2) dy$. The outer radius is $R(y)=2$ and the inner radius is $r(y)=\\sqrt{y}$. The bounds are from $y=0$ to $y=4$. The integral is $V = \\pi \int_{0}^{4} (2^2 - (\\sqrt{y})^2) dy = \\pi \int_{0}^{4} (4 - y) dy$. Evaluating the integral gives: $V = \\pi [4y - \\frac{y^2}{2}]_{0}^{4} = \\pi ((4(4) - \\frac{4^2}{2}) - 0) = \\pi (16 - 8) = 8\\pi$.

Correct Answer: B

The definite integral for volume is the accumulation of the areas of infinitesimally thin cross-sections. In the washer method, each cross-section is a ring or washer. The area of a circle is $\\pi r^2$. The area of a washer is the area of the larger circle minus the area of the smaller, inner circle (the hole). Therefore, $\\pi R(x)^2 - \\pi r(x)^2 = \\pi (R(x)^2 - r(x)^2)$ represents the area of one such ring-shaped cross-section.

Correct Answer: C

First, find the intersection points: $x = 2\\sqrt{x} \\Rightarrow x^2 = 4x \\Rightarrow x^2 - 4x = 0 \\Rightarrow x(x-4)=0$, so they intersect at $x=0$ and $x=4$. In the interval $(0, 4)$, the graph of $y=2\\sqrt{x}$ is above the graph of $y=x$. For revolution about the $x$-axis, the outer radius is $R(x) = 2\\sqrt{x}$ and the inner radius is $r(x) = x$. The volume is given by the washer method: $V = \\pi \int_{0}^{4} (R(x)^2 - r(x)^2) dx = \\pi \int_{0}^{4} ((2\\sqrt{x})^2 - x^2) dx = \\pi \int_{0}^{4} (4x - x^2) dx$. Option A represents the common error of squaring the difference of the radii instead of taking the difference of the squares.

Correct Answer: B

For revolution about the $y$-axis, we integrate with respect to $y$. The region's bounds in terms of $y$ are from $y=0$ (the x-axis) to $y=1$ (since $\\ln(e)=1$). The cross-sections are washers. The outer radius $R(y)$ is the distance from the $y$-axis to the line $x=e$, so $R(y)=e$. The inner radius $r(y)$ is the distance from the $y$-axis to the curve $y=\\ln x$, which must be rewritten as $x=e^y$, so $r(y)=e^y$. The volume is $V = \\pi \int_{0}^{1} (R(y)^2 - r(y)^2) dy = \\pi \int_{0}^{1} (e^2 - (e^y)^2) dy = \\pi \int_{0}^{1} (e^2 - e^{2y}) dy$.