A region is enclosed by the x-axis, the line
𝑥
=
1
, the line
𝑥
=
9
, and the curve
𝑦
=
𝑥
.

What is the volume of the solid generated when this region is rotated around the
𝑥
-axis?

Question

A region is enclosed by the x-axis, the line 
𝑥
=
1
, the line 
𝑥
=
9
, and the curve 
𝑦
=
𝑥
.

What is the volume of the solid generated when this region is rotated around the 
𝑥
-axis?

BlackTom AI · Accepted Answer

The problem describes a region bounded by the x-axis (y = 0), x = 1, x = 9, and the curve y = x, which lies above the x-axis in this interval. When this region is revolved around the x-axis, we can use the disk method: the radius of a typical disk at position x is y = x, so the cross-sectional area is π(y^2) = πx^2. Integrating from x = 1 to x = 9 gives the volume.

Compute the integral: V = ∫ from 1 to 9 of πx^2 dx = π [x^3/3] from 1 to 9 = π (9^3/3 − 1^3/3) = π (729/3 − 1/3) = π (728/3) = (728/3)π.

Now compare the given options to this exact value:
- Option A: 40π. This would correspond to 40π, which equals (120/3)π, far from (728/3)π. It underestimates the true volume by a large margin.
- Option B: 52 3 π. Interpreting as 52/3 π (approximately 17.33π) or as 52 3 π due to formatting, either way it does not match (728/3)π; it is far too small.
- Option C: None of these. Since the exact volume is (728/3)π, none of the listed numerical values equal this result.
- Option D: 81 2 π. Interpreting as 81/2 π (40.5π) or in any common misreading, it still does not equal (728/3)π; it does not match either.

In summary, the computed volume (728/3)π does not coincide with any of the provided numeric options, so the correct interpretation among the choices is that none of the listed values match the exact volume.

A region is enclosed by the x-axis, the line 𝑥 = 1 , the line 𝑥 = 9 , and the curve 𝑦 = 𝑥 . What is the volume of the solid generated when this region is rotated around the 𝑥 -axis?单项选择题

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