Disc Method: Revolving Around x/y Axes - AP Calculus BC

1. The region bounded by $x=2y+1$, $x=0$, and $0\le y\le 2$ is revolved about the y-axis; which integral is correct?
   Answer: $\pi\int_{0}^{2} (2y+1)^2\,dy$ This problem requires the disc method to find the volume of a solid of revolution. The region is between x = 2y + 1 and the y-axis (x=0) from y=0 to 2, revolving about the y-axis to form a solid without holes, so discs are used instead of washers. The radius is 2y + 1, the x-distance from the axis to the line. Squaring gives (2y + 1)², integrated with respect to y from 0 to 2. A tempting distractor like choice D uses x-variable and 2x + 1, failing to align with the y-axis revolution. Use the disc method when revolving a region between a curve and the axis of rotation with no gap creating a hole; opt for washers if there's an inner and outer radius.
2. The region between $y=\sqrt{x}$ and the $x$-axis from $x=0$ to $x=4$ is revolved about the $x$-axis; choose the volume setup.
   Answer: $\pi\int_{0}^{4}(\sqrt{x})^2\,dx$ This problem involves the disc method for finding volumes of solids of revolution. The disc method is appropriate since the region is between $y = \sqrt{x}$ and the $x$-axis, revolving around the $x$-axis, forming a solid with no holes. The radius of each disc is the function value $y = \sqrt{x}$ at each $x$. The volume is given by $\pi \int_{0}^{4} (\sqrt{x})^2 \, dx$, which simplifies to $\pi \int_{0}^{4} x \, dx$. Choice A is tempting as it matches the simplified integral, but it fails to show the explicit squaring of the radius, missing the conceptual setup. Remember, use the disc method when revolving a region directly against the axis of rotation without any gap; switch to washers if there's a hole in the solid.
3. What is the correct disc-method setup for revolving the region under $y=2x+1$ from $x=0$ to $x=2$ about the $x$-axis?
   Answer: $\pi\displaystyle\int_{0}^{2} (2x+1)^2\,dx$ This problem applies the disc method for rotation about the x-axis. When revolving the region under y = 2x + 1 from x = 0 to x = 2 about the x-axis, we create discs perpendicular to the x-axis. The radius of each disc is the y-value, which equals 2x + 1, and this radius must be squared in the disc formula: V = π∫₀² [2x + 1]² dx. Choice A incorrectly uses (2x + 1) without squaring, which would give the wrong volume formula. Choice D incorrectly includes a factor of 2π, which would be for the shell method, not discs. For disc method around the x-axis, use V = π∫[f(x)]² dx, not 2π∫f(x) dx.
4. Region bounded by $y=2x$, $y=0$, and $x=1$ is revolved about the x-axis; which integral sets up the volume?
   Answer: $\pi\int_{0}^{1} (2x)^2\,dx$ This problem requires the disc method to find the volume of a solid of revolution. The region is bounded by y = 2x and the x-axis, and revolving it about the x-axis creates a solid without holes, justifying discs over washers. The radius is the y-value from the axis to the curve, which is 2x. Squaring this gives (2x)², and we integrate with respect to x from 0 to 1, forming π ∫ from 0 to 1 of (2x)² dx. A tempting distractor like choice D introduces a 2π, which is for shells, but that's not suitable for this axis of rotation. Use the disc method when revolving a region between a curve and the axis of rotation with no gap creating a hole; opt for washers if there's an inner and outer radius.
5. A region bounded by $y=1+x^3$ and the $x$-axis for $-1\le x\le 0$ is revolved about the $x$-axis; which integral gives volume?
   Answer: $\pi\int_{-1}^{0}(1+x^3)^2\,dx$ This problem involves the disc method for finding volumes of solids of revolution. The disc method is appropriate as the region is between y = 1 + x³ and the x-axis from -1 to 0, revolving around the x-axis without holes. The radius is y = 1 + x³ at each x. The volume is π∫(1 + x³)² dx from -1 to 0. Choice A, π∫(1 + x³) dx, is a common distractor but fails by not squaring the radius, computing area instead. Remember, use the disc method when revolving a region directly against the axis of rotation without any gap; switch to washers if there's a hole in the solid.
6. The region between $y=\sqrt{x}$ and the x-axis for $0\le x\le 4$ is revolved about the x-axis; which volume integral is correct?
   Answer: $\pi\int_{0}^{4} (\sqrt{x})^2\,dx$ This problem uses the disc method since we're revolving the region between $y = \sqrt{x}$ and the x-axis around the x-axis, creating solid discs. The radius of each disc at position x is $R(x) = \sqrt{x}$, extending from the x-axis up to the curve. The area of each disc is $\pi [R(x)]^2 = \pi (\sqrt{x})^2 = \pi x$, and integrating from x = 0 to x = 4 gives the volume $\pi \int_0^4 (\sqrt{x})^2 \, dx$. Choice A incorrectly omits the $\pi$ factor and doesn't square the radius. Remember: disc method always requires squaring the radius function and including the $\pi$ factor.
7. The region bounded by $y=4-x$ and the $x$-axis on $0\le x\le 4$ is revolved about the $x$-axis; select the correct volume integral.
   Answer: $\pi\int_{0}^{4}(4-x)^2\,dx$ This problem involves the disc method for finding volumes of solids of revolution. Discs apply here as the region is bounded by y = 4 - x and the x-axis, revolved about the x-axis, creating a solid without holes. The radius is the height from the x-axis to the line, y = 4 - x. Integrate π times the squared radius along x from 0 to 4, giving π∫(4 - x)² dx. Choice A, π∫(4 - x) dx, is distracting but wrong because it skips squaring the radius, resembling an area calculation. Remember, use the disc method when revolving a region directly against the axis of rotation without any gap; switch to washers if there's a hole in the solid.
8. A region bounded by $y=\tan x$ and the x-axis on $0\le x\le \frac{\pi}{4}$ is revolved about the x-axis; choose the volume setup.
   Answer: $\pi\int_{0}^{\pi/4} (\tan x)^2\,dx$ This problem applies the disc method when the region bounded by y = tan x and the x-axis on 0 ≤ x ≤ π/4 is revolved about the x-axis. Since the region is solid (bounded by the curve and the axis of rotation), we form discs with radius R(x) = tan x. The volume is π∫₀^(π/4) [R(x)]² dx = π∫₀^(π/4) (tan x)² dx. Choice A omits the essential squaring of the radius, which would yield area instead of volume. The core rule: disc method about the x-axis always involves π times the y-function squared.
9. Revolve the region bounded by $x=|y|$ and the y-axis for $-2 \le y \le 2$ about the y-axis; choose the correct integral.
   Answer: $\pi\int_{-2}^{2} (|y|)^2\,dy$ This problem uses the disc method when the region bounded by $x = |y|$ and the y-axis for $-2 \le y \le 2$ is revolved about the y-axis. Since we have solid discs with radius $R(y) = |y|$ extending from the y-axis to the curve, the volume is $\pi \int_{-2}^{2} [R(y)]^2 \, dy = \pi \int_{-2}^{2} (|y|)^2 \, dy = \pi \int_{-2}^{2} y^2 \, dy$ (since $|y|^2 = y^2$). Choice A incorrectly omits the squaring of the radius function, giving area rather than volume. The fundamental principle: disc method about the y-axis always involves $\pi$ times the square of the x-function.
10. Revolve the region bounded by $y=\frac{x}{2}$ and the x-axis from $x=0$ to $x=4$ about the x-axis; select the correct integral.
    Answer: $\pi\int_{0}^{4} \left(\frac{x}{2}\right)^2\,dx$ This problem uses the disc method since the region bounded by y = x/2 and the x-axis from x = 0 to x = 4 is revolved about the x-axis, creating solid discs. The radius of each disc at position x is R(x) = x/2, extending from the x-axis to the curve. The volume integral is π∫₀⁴ [R(x)]² dx = π∫₀⁴ (x/2)² dx. Choice B incorrectly omits the squaring of the radius function, giving area under the curve rather than volume of revolution. The fundamental principle: disc method about the x-axis always requires π times the square of the y-function.
11. A region bounded by $x=\sqrt{1-y^2}$ and the y-axis on $-1 \le y \le 1$ is revolved about the y-axis; choose the setup.
    Answer: $\pi\int_{-1}^{1} (\sqrt{1-y^2})^2\,dy$ This problem uses the disc method when the region bounded by $x = \sqrt{1-y^2}$ and the y-axis on $-1 \leq y \leq 1$ is revolved about the y-axis. Since we have solid discs with radius $R(y) = \sqrt{1-y^2}$ extending from the y-axis to the curve, the volume is $\pi \int_{-1}^{1} [R(y)]^2 \, dy = \pi \int_{-1}^{1} [\sqrt{1-y^2}]^2 \, dy = \pi \int_{-1}^{1} (1-y^2) \, dy$. Choice B omits the squaring operation, and Choice E incorrectly squares the already-squared expression. The key principle: disc method requires squaring the radius function, so $[\sqrt{1-y^2}]^2 = 1-y^2$.
12. The region bounded by $y=\ln x$, $y=0$, and $x=e$ is revolved about the $x$-axis; which integral gives the volume?
    Answer: $\pi\displaystyle\int_{1}^{e} (\ln x)^2\,dx$ This problem requires the disc method for revolution around the x-axis. We're revolving the region bounded by y = ln x, y = 0, and x = e around the x-axis. Since ln x is only defined for x > 0 and ln(1) = 0, the region extends from x = 1 to x = e, not from x = 0. The radius of each disc is |ln x|, and since ln x ≥ 0 on [1, e], the area is π(ln x)². Choice A incorrectly uses 0 as the lower limit, but ln x is undefined at x = 0. When setting up disc method integrals, always verify that the function is defined over the entire interval of integration.
13. Region bounded by $y=\ln x$, the $x$-axis, and $x=1$ to $x=e$ is revolved about the $x$-axis; choose the setup.
    Answer: $\pi\int_{1}^{e} (\ln x)^2\,dx$ This problem applies the disc method for rotation about the x-axis. The region is bounded above by y = ln x and below by the x-axis, from x = 1 to x = e (since ln(1) = 0 and ln(e) = 1). When revolving about the x-axis, each disc has radius R(x) = y = ln x, so the disc area is π[R(x)]² = π(ln x)². The volume integral becomes π∫₁ᵉ (ln x)² dx. Choice B incorrectly uses π ln x without squaring, while choice C has impossible limits since ln x is undefined at x = 0. For disc method, always square the radius function to get the disc area.
14. Revolve the region under $y=\frac{1}{2}x+1$ above the x-axis from $x=0$ to $x=2$ about the x-axis; choose the integral.
    Answer: $\pi\int_{0}^{2} (\tfrac{1}{2}x+1)^2\,dx$ This problem uses the disc method for revolving the region under $y = \frac{1}{2}x + 1$ above the x-axis from $x = 0$ to $x = 2$ about the x-axis. The region creates solid discs with radius $R(x) = \frac{1}{2}x + 1$ extending from the axis of rotation to the curve. Each disc has area $\pi [R(x)]^2 = \pi \left( \frac{1}{2}x + 1 \right)^2$, giving volume $\pi \int_0^2 \left( \frac{1}{2}x + 1 \right)^2 \, dx$. Choice B incorrectly omits the squaring of the radius function, calculating area instead of volume. The fundamental principle: disc method about the x-axis always requires $\pi$ times the square of the y-function.
15. Revolve the region bounded by $x=\frac{3}{2}y$ and the y-axis for $0\le y\le 2$ about the y-axis; choose the correct setup.
    Answer: $\pi\int_{0}^{2} \left(\frac{3}{2}y\right)^2\,dy$ This problem uses the disc method when the region bounded by x = (3/2)y and the y-axis for 0 ≤ y ≤ 2 is revolved about the y-axis. Since we're revolving around the y-axis, the radius at height y is R(y) = (3/2)y, extending from the y-axis to the curve. The volume is π∫₀² [R(y)]² dy = π∫₀² [(3/2)y]² dy. Choice B incorrectly omits the squaring of the radius function, yielding area instead of volume. The essential principle: disc method about the y-axis always involves π times the square of the x-function.
16. The region bounded by $y=2x-x^2$ and the x-axis on $0\le x\le 2$ is revolved about the x-axis; which integral gives volume?
    Answer: $\pi\int_{0}^{2} (2x-x^2)^2\,dx$ This problem applies the disc method since the region bounded by y = 2x - x² and the x-axis on 0 ≤ x ≤ 2 is revolved about the x-axis, forming solid discs. The radius of each disc is R(x) = 2x - x², extending from the x-axis up to the curve. The volume integral is π∫₀² [R(x)]² dx = π∫₀² (2x - x²)² dx. Choice B omits the crucial squaring of the radius function, yielding area under the curve rather than volume of revolution. The fundamental rule: disc method about the x-axis always requires π times the square of the y-function.
17. The region bounded by $x=4-y^2$ and the y-axis on $-2\le y\le 2$ is revolved about the y-axis; which integral gives volume?
    Answer: $\pi\int_{-2}^{2} (4-y^2)^2\,dy$ This problem uses the disc method since the region bounded by x = 4 - y² and the y-axis on -2 ≤ y ≤ 2 is revolved about the y-axis, creating solid discs. The radius of each disc at height y is R(y) = 4 - y², extending from the y-axis to the curve. The volume integral is π∫₋₂² [R(y)]² dy = π∫₋₂² (4 - y²)² dy. Choice B incorrectly omits the squaring of the radius function, giving area instead of volume. The essential principle: disc method about the y-axis always involves π times the square of the x-function for proper volume calculation.
18. Which integral represents the volume when the region under $y=3- x$ from $x=0$ to $x=3$ is revolved about the $x$-axis?
    Answer: $\pi\displaystyle\int_{0}^{3} (3-x)^2\,dx$ This problem uses the disc method for revolving around the x-axis. When the region under y = 3 - x from x = 0 to x = 3 is revolved about the x-axis, we form circular discs perpendicular to the x-axis. The radius of each disc equals the y-value, which is 3 - x, so we square this radius in the disc formula: V = π∫₀³ (3 - x)² dx. Choice B incorrectly uses (3 - x) without squaring the radius function. Choice D uses 2π∫ (3 - x) dx, which would be the shell method formula, not the disc method. When using discs for rotation about the x-axis, the integrand must contain the square of the radius function.
19. A region bounded by $y=\frac{1}{2}(x-1)$ and the x-axis on $1\le x\le 3$ is revolved about the x-axis; select the volume integral.
    Answer: $\pi\int_{1}^{3} \left(\frac{x-1}{2}\right)^2\,dx$ This problem applies the disc method since the region bounded by $y = \frac{1}{2}(x-1)$ and the x-axis on $1 \leq x \leq 3$ is revolved about the x-axis, forming solid discs. The radius of each disc at position x is $R(x) = \frac{1}{2}(x-1)$, extending from the x-axis to the curve. The volume integral is $\pi\int_1^3 [R(x)]^2 \, dx = \pi\int_1^3 \left( \frac{x-1}{2} \right)^2 \, dx$. Choice B omits the essential squaring of the radius, which would give area rather than volume. The core rule: disc method about the x-axis requires $\pi$ times the y-function squared to calculate volume correctly.
20. Revolve the region bounded by $x=\sqrt{y+1}$ and the y-axis for $-1\le y\le 3$ about the y-axis; choose the correct setup.
    Answer: $\pi\int_{-1}^{3} (\sqrt{y+1})^2\,dy$ This problem uses the disc method since the region bounded by x = √(y+1) and the y-axis for -1 ≤ y ≤ 3 is revolved about the y-axis, creating solid discs. The radius of each disc at height y is R(y) = √(y+1), extending from the y-axis to the curve. The volume integral is π∫₋₁³ [R(y)]² dy = π∫₋₁³ [√(y+1)]² dy = π∫₋₁³ (y+1) dy. Choice A omits the squaring operation, and the essential principle is: disc method requires squaring the radius function, so [√(y+1)]² = y+1.

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