To find the volume of the solid generated by revolving the region bounded by $y = 11x - 10$, $y = \sqrt{x}$, and $x = 0$ about the y-axis using the shell method, we will follow these steps:

Step 1: Understanding the shell method

The formula for the volume of a solid generated by revolving a region about the y-axis using the shell method is:

$V = 2\pi \int_{a}^{b} \text{(radius of shell)} \times \text{(height of shell)} \, dx$

• The radius of the shell is the distance from the y-axis, which is just $x$.
• The height of the shell is the difference between the upper curve $y = 11x - 10$ and the lower curve $y = \sqrt{x}$ at a particular value of $x$.

Step 2: Determine the bounds of integration

To determine the bounds of integration, we need to find the points of intersection of $y = 11x - 10$ and $y = \sqrt{x}$.

Set $11x - 10 = \sqrt{x}$:

$11x - 10 = \sqrt{x}$

Squaring both sides to eliminate the square root:

$(11x - 10)^2 = x$

Expanding the left-hand side:

$121x^2 - 220x + 100 = x$

Simplifying:

$121x^2 - 221x + 100 = 0$

Now, we solve this quadratic equation using the quadratic formula:

$x = \frac{-(-221) \pm \sqrt{(-221)^2 - 4(121)(100)}}{2(121)}$ $x = \frac{221 \pm \sqrt{48841 - 48400}}{242}$ $x = \frac{221 \pm \sqrt{441}}{242}$ $x = \frac{221 \pm 21}{242}$

Thus, the two solutions are:

$x = \frac{221 + 21}{242} = \frac{242}{242} = 1$ $x = \frac{221 - 21}{242} = \frac{200}{242} = \frac{100}{121}$

Therefore, the points of intersection are at $x = 1$ and $x = \frac{100}{121}$. These will be the bounds for the integration.

Step 3: Set up the integral for the volume

The radius of a shell is $x$, and the height of the shell is the difference between the two curves, $(11x - 10) - \sqrt{x}$.

The volume integral is:

$V = 2\pi \int_{\frac{100}{121}}^{1} x \left( (11x - 10) - \sqrt{x} \right) dx$

Step 4: Simplify and compute the integral

We now expand the integrand:

$V = 2\pi \int_{\frac{100}{121}}^{1} \left( x(11x - 10) - x\sqrt{x} \right) dx$ $V = 2\pi \int_{\frac{100}{121}}^{1} \left( 11x^2 - 10x - x^{3/2} \right) dx$

Now, integrate each term separately:

$\int 11x^2 \, dx = \frac{11x^3}{3}$ $\int -10x \, dx = -5x^2$ $\int -x^{3/2} \, dx = -\frac{2x^{5/2}}{5}$

Thus, the volume is:

$V = 2\pi \left[ \frac{11x^3}{3} - 5x^2 - \frac{2x^{5/2}}{5} \right]_{\frac{100}{121}}^{1}$

Step 5: Evaluate the integral

Substitute the bounds $x = 1$ and $x = \frac{100}{121}$ into the expression for the volume.

At $x = 1$: $\frac{11(1)^3}{3} - 5(1)^2 - \frac{2(1)^{5/2}}{5} = \frac{11}{3} - 5 - \frac{2}{5}$

At $x = \frac{100}{121}$, plug in $x = \frac{100}{121}$ and simplify.

Finally, subtract the two values and multiply by $2\pi$.

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