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 where the curves meet.
Setting $x^2+2=4x-1$ gives $x^2-4x+3=0$, so $x=1$ and $x=3$.

Step 2: Decide top and bottom.
Between $1$ and $3$ the line $4x-1$ is above the parabola $x^2+2$, so the line is the outer radius.

Step 3: Use the washer formula.
\[ V=\pi\int_1^3\left[(4x-1)^2-(x^2+2)^2\right]dx=\pi\int_1^3(-x^4+12x^2-8x-3)\,dx. \]

Step 4: Integrate.
The antiderivative is $-\frac{x^5}5+4x^3-4x^2-3x$. At $x=3$ it is $\frac{72}5$, at $x=1$ it is $-\frac{16}5$. The difference is $\frac{88}5$.

Step 5: Conclude.
\[ \boxed{\frac{88}{5}\pi} \]