This problem uses the disc method for revolution about the x-axis. Since the region is bounded by $y = 2x - x^2$ and the x-axis with no inner boundary, discs are the appropriate method. The radius of each disc at position x is the distance from the x-axis to the curve: $r(x) = 2x - x^2$. The volume formula becomes $V = \pi \int_a^b [r(x)]^2 , dx = \pi \int_0^2 (2x - x^2)^2 , dx$. Choice B fails because it doesn't square the radius function, which is essential in the disc method. Apply the disc method when there's a single boundary curve extending from the axis of revolution.

This problem uses the disc method for revolution about the x-axis. Since the region is bounded by y = ln(x) and the x-axis with no inner boundary, discs rather than washers apply. The radius of each disc at position x is r(x) = ln(x), the distance from the x-axis to the curve. The volume is V = π∫[a to b] [r(x)]² dx = π∫[1 to e] (ln x)² dx. Choice B fails because it doesn't square the radius function, which is essential in the disc method formula. Use discs when the solid extends from the axis of revolution to a single bounding curve.

This problem uses the disc method for revolution about the y-axis. Since the region is bounded by x = cos y and the y-axis with no inner boundary, discs are the appropriate approach. The radius of each disc at position y is r(y) = cos y, the distance from the y-axis to the curve. The volume is V = π∫[a to b] [r(y)]² dy = π∫[0 to π/2] (cos y)² dy. Choice A fails because it doesn't square the radius function, which is essential in the disc method. Apply discs when there's exactly one boundary curve extending from the axis of revolution.

This problem applies the disc method for finding volume when revolving about the x-axis. Since the region is bounded by $y = 3x$ and the x-axis with no inner boundary, we use discs rather than washers. The radius of each disc at position x is the distance from the x-axis to the curve: $r(x) = 3x$. The volume is $V = \pi \int_{a}^{b} [r(x)]^2 , dx = \pi \int_{0}^{2} (3x)^2 , dx$. Choice B is tempting but incorrect because it omits the essential squaring of the radius in the disc method formula. Apply discs when the solid extends from the axis of revolution to exactly one bounding curve.

This problem uses the disc method for rotation about the x-axis. Revolving the region under y = ln x from x = 1 to x = e around the x-axis creates circular discs perpendicular to the x-axis. The radius of each disc equals the y-value, which is ln x, making the disc's area π(ln x)². We integrate from x = 1 to x = e (where ln(1) = 0 and ln(e) = 1) to find the volume: π∫₁ᵉ(ln x)²dx. Choice B incorrectly uses π∫₁ᵉ ln x dx, missing the essential squaring of the radius needed for disc area calculation. Remember: disc method around the x-axis requires π times the square of the function value, which represents the area of each circular cross-section.

This problem uses the disc method for revolving around the x-axis. When rotating the region under y = 1 + x³ from x = 0 to x = 1 around the x-axis, we create circular discs perpendicular to the x-axis. Each disc has radius equal to the y-value at that x-position, which is 1 + x³, so the disc's area is π(1 + x³)². The volume is the integral of these disc areas: π∫₀¹(1 + x³)²dx. Choice B incorrectly uses π∫₀¹(x³)²dx, which would only work if the function were y = x³ instead of y = 1 + x³—this error ignores the constant term that shifts the curve up by 1 unit. The disc method requires squaring the entire function value, not just part of it: use π∫[f(x)]²dx where f(x) is the complete function.

This problem uses the disc method for y = √(1 - x²) revolved around the x-axis. The function represents the upper semicircle of radius 1, and when revolving around the x-axis, each disc has radius √(1 - x²). The disc method requires V = π∫₀¹[√(1 - x²)]² dx = π∫₀¹(1 - x²) dx. The bounds are from x = 0 to x = 1, representing the first quadrant portion of the semicircle. Choice A incorrectly uses √(1 - x²) without squaring, which would give a linear measure rather than the disc's area. For disc method, always square the radius function—even when it contains a square root, the squaring simplifies the expression.

This problem applies the disc method for revolution about the x-axis. The region bounded by y = x/(x+1) and the x-axis creates a solid with no hole, so discs rather than washers are appropriate. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = x/(x+1). Using the disc method formula: $V = \pi \int_a^b [r(x)]^2 , dx = \pi \int_0^2 \left( \frac{x}{x+1} \right)^2 , dx$. Choice B is incorrect because it omits the required squaring of the radius in the disc method formula. Use discs when the solid extends from the axis of revolution to a single boundary curve.

This problem applies the disc method for revolution about the x-axis. The region bounded by y = tan x and the x-axis creates a solid with no hole, so discs rather than washers are appropriate. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = tan x. Using the disc method formula: V = π∫[a to b] [r(x)]² dx = π∫[0 to π/4] (tan x)² dx. Choice B is incorrect because it omits the required squaring of the radius in the disc method formula. Use discs when the solid extends from the axis of revolution to a single boundary curve.

This problem applies the disc method for revolution about the x-axis. The region bounded by y = 1/(1+x²) and the x-axis creates a solid with no hole, so discs rather than washers are appropriate. The radius of each disc at position x equals the distance from the x-axis to the curve: r(x) = 1/(1+x²). Using the disc method formula: V = π∫[a to b] [r(x)]² dx = π∫[0 to 1] (1/(1+x²))² dx. Choice B is incorrect because it omits the required squaring of the radius in the disc method formula. Use discs when the solid extends from the axis of revolution to a single boundary curve.