Question

The bounded region enclosed by the curve \(y=x^2+2\) and the line \(y=4x-1\) is revolved about the \(x\)-axis to generate a solid. The volume of the generated solid equals

• A. \(\frac{88}{5}\pi\)
• B. \(\frac{84}{5}\pi\)
• C. \(\frac{82}{5}\pi\)
• D. \(\frac{92}{5}\pi\)

Hint:

When a region between two curves is revolved about the \(x\)-axis, use the washer method: \(V=\pi\int_a^b(R^2-r^2)\,dx\), where \(R\) is the upper curve and \(r\) is the lower curve.

Answer 1

The Correct Option is A

Step 1: Find the points of intersection.
The given curves are
\[ y=x^2+2 \] and
\[ y=4x-1 \] Equating them,
\[ x^2+2=4x-1 \] \[ x^2-4x+3=0 \] \[ (x-1)(x-3)=0 \] Therefore,
\[ x=1,3 \] So, the bounded region lies between \(x=1\) and \(x=3\).

Step 2: Identify outer and inner radius.
For \(1<x<3\), the line \(y=4x-1\) lies above the curve \(y=x^2+2\).
Hence, when revolved about the \(x\)-axis,
\[ R=4x-1 \] and
\[ r=x^2+2 \]

Step 3: Use washer method.
The volume is
\[ V=\pi\int_{1}^{3}\left(R^2-r^2\right)\,dx \] \[ V=\pi\int_{1}^{3}\left[(4x-1)^2-(x^2+2)^2\right]\,dx \] Now,
\[ (4x-1)^2=16x^2-8x+1 \] and
\[ (x^2+2)^2=x^4+4x^2+4 \] Therefore,
\[ (4x-1)^2-(x^2+2)^2 = -x^4+12x^2-8x-3 \] So,
\[ V=\pi\int_{1}^{3}\left(-x^4+12x^2-8x-3\right)\,dx \]

Step 4: Evaluate the integral.
\[ V=\pi\left[-\frac{x^5}{5}+4x^3-4x^2-3x\right]_{1}^{3} \] At \(x=3\),
\[ -\frac{3^5}{5}+4(3^3)-4(3^2)-3(3) = -\frac{243}{5}+108-36-9 \] \[ = -\frac{243}{5}+63 = \frac{72}{5} \] At \(x=1\),
\[ -\frac{1}{5}+4-4-3 = -\frac{1}{5}-3 = -\frac{16}{5} \] Therefore,
\[ V=\pi\left[\frac{72}{5}-\left(-\frac{16}{5}\right)\right] \] \[ V=\pi\cdot \frac{88}{5} \] \[ V=\frac{88}{5}\pi \]

Step 5: Final conclusion.
Hence, the volume of the generated solid is
\[ \boxed{\frac{88}{5}\pi} \]