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.

MAT137Y1 LEC 20249: Calculus with Proofs (all lecture sections) Pre-Class Quiz 48 (10.1)

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?

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