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?

View Explanation

Log in for full answers

Similar Questions

The area bounded by the curve \( y=2sin(2x) \), the y-axis and the line y = 2 is rotated about the x-axis. The volume formed is equal to:

The area bounded by the curve y=2sin(2x)[math] y=2sin(2x) , the y-axis and the line y = 2 is rotated about the x-axis. The volume formed is equal to:

Question textThe volume of the solid of revolution formed by rotating the curve \(y=\text{Arcsin}(\frac{x}{2}), 0\leq x\leq 2\) about the \(y\)-axis is Answer 1 Question 34[input]\(\pi^2\).

Question textThe volume generated by rotating, about the \(X\) axis, the region enclosed by \(y=x^{\frac{3}{2}}\), \(x=1,x=2\), and the \(X\) axis, is Answer 1 Question 5[input] \(\pi \big/\) Answer 2 Question 5[input].

Question textThe volume of the solid of revolution formed by rotating the curve \(y=\text{Arcsin}(\frac{x}{2}), 0\leq x\leq 2\) about the \(y\)-axis is Answer 1 Question 4[input]\(\pi^2\).

Question textThe volume generated by rotating, about the \(X\) axis, the region enclosed by \(y=x^2\), \(x=1,x=2\), and the \(X\) axis, is Answer 1 Question 2[input] \(\pi \big/\) Answer 2 Question 2[input].

A solid of revolution is formed by rotating the region between the graph of g(y)= y3 3 and the y-axis over 1<y<2, around the y -axis. Which of the following integrals give the volume of the solid?

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler