Problem 35 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by \(y=x^{2}, x=1,\) and \(y=0 .\) Use the shell method to find the volume of the solid generated when \(R\) is revolved about the following lines. $$y=-2$$

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{28\pi}{3}\) cubic units.

Step 1: Determine the height of the bounding curve

Since the region is bounded by \(y=0\), and we are revolving around \(y=-2\), the height of the bounding curve at a given \(x\) is simply \(-2 - 0 = -2\). So, \(h=-2\).

Step 2: Determine the radial distance

The radial distance at any given point is the distance from the horizontal line \(y=-2\) to the bounding curve \(y=x^2\). To find this distance, write down the equation of the line connecting \((x, x^2)\) to \((x, -2)\). To find the length of the line, you can simply subtract the \(y\)-coordinate of the points \((x^2 - (-2))\):$$f(x) = x^2 - (-2) = x^2 + 2$$.

Step 3: Apply the shell method formula

Plug the values we found into the shell method formula for revolving around a horizontal line:$$V = 2\pi\int_0^1 (-2)(x^2 + 2) dx$$

Step 4: Evaluate the integral

To find the volume, we need to evaluate the integral:$$V = 2\pi \left[-2 \int_0^1 x^2 dx - 4 \int_0^1 dx \right]$$Evaluate the two integrals separately:$$\int_0^1 x^2 dx = \frac{x^3}{3}\Big|_0^1 = \frac{1}{3}$$$$\int_0^1 dx = x\Big|_0^1 = 1$$Plug these results back into the volume equation:$$V = 2\pi \left[-2\left(\frac{1}{3}\right) - 4(1) \right]$$

Step 5: Simplify and find the volume

Simplify the expression:$$V = 2\pi \left[-\frac{2}{3} - 4 \right] = 2\pi \left[-\frac{14}{3} \right] = -\frac{28\pi}{3}$$Since the volume cannot be negative, take the absolute value:$$V = \frac{28\pi}{3}$$The volume of the solid generated when the region bounded by \(y=x^2, x=1,\) and \(y=0\) is revolved about the line \(y=-2\) is \(\frac{28\pi}{3}\) cubic units.