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].

Question

BlackTom AI · Accepted Answer

We are asked for the volume generated by rotating the region around the X-axis that is bounded by y = x^2, x = 1, x = 2, and the X-axis.
First, visualize the setup: the region lies between x = 1 and x = 2 and above the X-axis since y = x^2 is nonnegative in this interval.
When rotating about the X-axis, the disc (washer) method gives the volume V = π ∫_{a}^{b} [radius(x)]^2 dx, where the radius is the distance from the X-axis to the curve, i.e., radius = y = x^2.
Thus the integrand is [radius]^2 = (x^2)^2 = x^4, and the limits are from x = 1 to x = 2.
Compute the integral: ∫_{1}^{2} x^4 dx = [x^5/5]_{1}^{2} = (2^5)/5 − (1^5)/5 = 32/5 − 1/5 = 31/5.
Multiply by π for the volume: V = π × (31/5) = (31π)/5.
Now match the two blanks in the prompt: the first blank corresponds to the numerator of the fraction inside the parentheses, which is 31; the second blank corresponds to the denominator, which is 5.
So the completed form is π × (31/5) or equivalently (31π)/5, and the filled numbers are 31 and 5 respectively.
If someone mistakenly used the boundary y = x^2 as the radius without squaring or used wrong limits, they might get an incorrect coefficient, but here the proper setup and computation lead directly to the 31/5 factor times π.
In summary, the first blank should be 31, and the second blank should be 5.

ENG1090 - MUM S1 2025 Lecture quiz 11

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].

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]\(\pi^2\).

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?

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