This problem applies the disc method since the region bounded by $x = e^y$ and the y-axis for $0 \le y \le 1$ is revolved about the y-axis, forming solid discs. The radius of each disc at height y is $R(y) = e^y$, extending from the y-axis to the curve. The volume integral is $\pi \int_0^1 [R(y)]^2 , dy = \pi \int_0^1 (e^y)^2 , dy$. Choice A omits the essential squaring of the radius, which would yield area instead of volume. The core rule for disc method about the y-axis: always square the x-function and multiply by $\pi$.

This problem applies the disc method when the region bounded by x = y² 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) = y², extending from the y-axis to the curve. Each disc has area π[R(y)]² = π(y²)², and the volume integral is π∫₀² (y²)² dy. Choice B omits the crucial squaring of the radius, yielding area instead of volume. The essential rule: for disc method about the y-axis, square the x-function and integrate with respect to y.

This problem applies the disc method since the region under y = 1/(1-x) above the x-axis on 0 ≤ x ≤ 1/2 is revolved about the x-axis, forming solid discs. The radius of each disc at position x is R(x) = 1/(1-x), extending from the x-axis to the curve. The volume integral is π∫₀^(1/2) [R(x)]² dx = π∫₀^(1/2) [1/(1-x)]² dx. Choice A omits the crucial squaring of the radius function, yielding area under the curve rather than volume of revolution. The key rule: disc method about the x-axis always requires π times the square of the y-function.

This problem applies the disc method when the region between y = eˣ and the x-axis from x = 0 to x = 1 is revolved about the x-axis. The region creates solid discs with radius R(x) = eˣ extending from the axis of rotation to the curve. Each disc has area π[R(x)]² = π(eˣ)², giving volume π∫₀¹ (eˣ)² dx. Choice A omits the essential squaring of the radius, yielding area rather than volume. The core rule for disc method: always square the radius function and multiply by π to obtain volume.

This problem uses the disc method for revolving the region under y = x³ above the x-axis from x = 0 to x = 1 about the x-axis. The region creates solid discs with radius R(x) = x³ extending from the axis of rotation to the curve. Each disc has area π[R(x)]² = π(x³)², giving volume π∫₀¹ (x³)² dx. Choice A incorrectly omits the squaring of the radius, which would calculate area instead of volume. The essential principle: disc method about the x-axis requires π times the square of the y-function for proper volume calculation.

This problem applies the disc method when the region bounded by y = √(x+1) and the x-axis on -1 ≤ x ≤ 3 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) = √(x+1). The volume is π∫₋₁³ [R(x)]² dx = π∫₋₁³ [√(x+1)]² dx = π∫₋₁³ (x+1) dx. Choice A omits the crucial squaring operation, and the key principle is: disc method requires squaring the radius function, so [√(x+1)]² = x+1.

This problem applies the disc method for revolving the region bounded by y = 1 - x² and the x-axis for -1 ≤ x ≤ 1 about the x-axis. The region forms solid discs with radius R(x) = 1 - x² extending from the axis of rotation to the curve. The volume integral is π∫₋₁¹ [R(x)]² dx = π∫₋₁¹ (1 - x²)² dx. Choice B omits the crucial squaring of the radius, which would calculate area under the curve rather than volume of revolution. The key rule: disc method requires π times the radius function squared.

This problem applies the disc method when the region bounded by x = 3y and the y-axis from y = 0 to y = 1 is revolved about the y-axis. Since the region is solid (bounded by the curve and the axis of rotation), we form discs with radius R(y) = 3y. The volume integral is π∫₀¹ [R(y)]² dy = π∫₀¹ (3y)² dy. Choice B omits the essential squaring of the radius, which would calculate area instead of volume. The core rule: disc method about the y-axis always requires π times the square of the x-function.

This problem involves the disc method for finding volumes of solids of revolution. The disc method applies as the region is between y = 2x and the x-axis from 0 to 5, revolved about the x-axis, with no holes. The radius is y = 2x at each x. The volume integral is π∫(2x)² dx = π∫4x² dx from 0 to 5. Choice B, π∫2x dx, is tempting but incorrect since it omits squaring the radius, treating it like area. 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.

This problem requires the disc method to find the volume of a solid of revolution. The region is bounded by the curve and the x-axis, and when revolved about the x-axis, it forms a solid without holes, making the disc method appropriate rather than washers. The radius of each disc is the distance from the x-axis to the curve, which is given by y = √x. To set up the integral, we square this radius and integrate with respect to x from 0 to 4, yielding π ∫ from 0 to 4 of $(√x)^2$ dx. A tempting distractor like choice D includes a 2π factor, which would apply to the shell method instead, but that's incorrect here since we're revolving around the x-axis using discs. 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.