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].

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BlackTom AI · Accepted Answer

The problem asks for the volume generated by revolving the region bounded by y = x^{3/2}, x = 1, x = 2, and the x-axis around the x-axis.
First, identify the radius for the disk method: at a given x between 1 and 2, the region is between the x-axis (y = 0) and the curve y = x^{3/2}, so the radius is y = x^{3/2}.
Using the disk method about the x-axis, the volume is V = π ∫_{1}^{2} [radius]^2 dx = π ∫_{1}^{2} (x^{3/2})^2 dx.
Compute the integrand: (x^{3/2})^2 = x^3.
Thus V = π ∫_{1}^{2} x^3 dx = π [x^4 / 4]_{1}^{2}.
Evaluate the antiderivative at the bounds: x = 2 gives 16/4 = 4, and x = 1 gives 1/4. The difference is 4 − 1/4 = 16/4 − 1/4 = 15/4.
Therefore V = π × (15/4) = (15/4) π.
In the entered blanks, this corresponds to the fraction 15/4 being placed next to π, i.e., the numerator 15 and the denominator 4 to complete the expression π · (15/4). 
Common potential mistakes include:
- Using the incorrect radius (e.g., forgetting to square the radius when applying the disk formula).
- Mixing up the bounds or miscomputing ∫ x^3 dx (which is x^4/4).
- Forgetting the π factor or misplacing it outside the integral.
- Not applying the correct setup for revolving around the x-axis (confusing with shells, which would lead to a different integral form).
Carefully performing each step shows the volume is (15/4)π, as obtained by evaluating the definite integral with the given limits.

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].Multiple fill-in-the-blank

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