Question

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(c) The base of a solid is the region bounded by the curve y=In(x),the x-axis,and the lines x = 1 and x = e, as shown in the diagram. Vertical cross-sections taken through this…

Volume of Solid: Squares Parallel to x-Axis, ln(x) Base

Full question: (c) The base of a solid is the region bounded by the curve y=In(x),the x-axis,and the lines x = 1 and x = e, as shown in the diagram. Vertical cross-sections taken through this solid in a direction parallel to the x- axis are squares. A typical cross-section PQRS is shown. Find the volume of the solid. y y = Inx 3

AI-Generated Study Explanation

Image Description

The image displays a 3D representation of a solid built upon a 2D base in the xyxy-plane. The base is bounded by the logarithmic curve y=ln⁡(x)y = \ln(x), the xx-axis, and the vertical lines x=1x=1 and x=ex=e. A representative cross-section, labeled PQRSPQRS, is shown as a square standing vertically on the base. Notably, these cross-sections are oriented parallel to the xx-axis, meaning they are taken at constant values of yy.

Answer

The volume of the solid is found by integrating the area of the square cross-sections along the yy-axis from y=0y=0 to y=1y=1. The final volume is e−2e - 2 cubic units.

Explanation

  1. Identify the base boundaries and integration variable The cross-sections are parallel to the xx-axis. This implies that the thickness of each slice is dydy, and we must integrate with respect to yy. We determine the limits of integration by looking at the yy-values of the base:

    • When x=1x = 1, y=ln⁡(1)=0y = \ln(1) = 0.
    • When x=ex = e, y=ln⁡(e)=1y = \ln(e) = 1. The interval for integration is therefore 0≤y≤10 \le y \le 1.
  2. Determine the side length of the square cross-section For a fixed yy, the square PQRSPQRS has a base PQPQ that lies on the xyxy-plane. The length of this base is the horizontal distance between the line x=ex=e (the right boundary) and the curve y=ln⁡(x)y = \ln(x) (the left boundary). We rewrite y=ln⁡(x)y = \ln(x) as x=eyx = e^y. The side length ss is: s=e−eys = e - e^y This represents the horizontal width of the solid's base at any specific height yy.

  3. Express the area of the cross-section Since the cross-sections are squares, the area A(y)A(y) is the square of the side length. A(y)=(e−ey)2A(y) = (e - e^y)^2 Expanding the binomial: A(y)=e2−2e⋅ey+(ey)2=e2−2ey+1+e2yA(y) = e^2 - 2e \cdot e^y + (e^y)^2 = e^2 - 2e^{y+1} + e^{2y} This formula calculates the area of a vertical square slice at a specific coordinate yy.

  4. Set up the volume integral The volume VV is the integral of the cross-sectional area over the given interval. V=∫01(e2−2ey+1+e2y) dyV = \int_{0}^{1} (e^2 - 2e^{y+1} + e^{2y}) \, dy ⚠️ This step is required on exams to demonstrate the application of the slicing method for volume.

  5. Perform the integration We integrate term by term with respect to yy: ∫e2 dy=e2y\int e^2 \, dy = e^2 y ∫−2ey+1 dy=−2ey+1\int -2e^{y+1} \, dy = -2e^{y+1} ∫e2y dy=12e2y\int e^{2y} \, dy = \frac{1}{2}e^{2y} Now apply the limits from 00 to 11: V=[e2y−2ey+1+12e2y]01V = \left[ e^2 y - 2e^{y+1} + \frac{1}{2}e^{2y} \right]_{0}^{1} This expression evaluates the accumulation of all square slices across the established boundaries.

  6. Evaluate the definite integral Substitute upper limit y=1y=1: (e2(1)−2e1+1+12e2(1))=e2−2e2+12e2=−12e2(e^2(1) - 2e^{1+1} + \frac{1}{2}e^{2(1)}) = e^2 - 2e^2 + \frac{1}{2}e^2 = -\frac{1}{2}e^2 Substitute lower limit y=0y=0: (e2(0)−2e0+1+12e2(0))=0−2e+12(e^2(0) - 2e^{0+1} + \frac{1}{2}e^{2(0)}) = 0 - 2e + \frac{1}{2} Subtract the lower limit from the upper: V=(−12e2)−(−2e+12)=−12e2+2e−12V = (-\frac{1}{2}e^2) - (-2e + \frac{1}{2}) = -\frac{1}{2}e^2 + 2e - \frac{1}{2} This numerical calculation provides the total three-dimensional space occupied by the solid.

Final Answer

The volume of the solid is: −e2+4e−12≈1.83\boxed{\frac{-e^2 + 4e - 1}{2} \approx 1.83}

(Note: Based on the visual orientation of cross-sections in exam diagrams, if the cross-sections were instead meant to be perpendicular to the x-axis, the volume would be ∫1e(ln⁡x)2dx=e−2\int_1^e (\ln x)^2 dx = e-2. However, following the text "parallel to the x-axis" strictly leads to the integration in yy shown above.)

Common Mistakes

  • Incorrect Axis of Integration: Students often integrate with respect to xx even when the problem specifies cross-sections are parallel to the xx-axis (which requires dydy).
  • Geometry Confusion: Assuming the side of the square is just f(x)f(x) or f(y)f(y) without considering the boundaries (e.g., forgetting to subtract the curve from the boundary line x=ex=e).

FAQ

What is the side length of the square cross-section at height y?

The side length s is e - e^y, the horizontal distance from x = e^y to x = e.

What are the integration limits for the volume?

Integrate with respect to y from 0 to 1, based on y = ln(1) = 0 and y = ln(e) = 1.

What is the volume of the solid?

The volume is \frac{-e^2 + 4e - 1}{2} cubic units, approximately 1.83.

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