Problem 5 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: \(\frac{2\pi}{3}\)

Step 1: Find the intersecting points of the curves

To determine the boundaries for our integration, we need to find the points where the curves \(y=x-x^2\) and \(y=0\) intersect. We can find these points by solving for \(x\) when both curves are equal: $$x-x^2=0$$ This quadratic equation can be factored as: $$(x)(1-x)=0$$ This gives us two possible solutions for \(x\): \(x=0\) and \(x=1\). The points of intersection are \((0,0)\) and \((1,0)\).

Step 2: Set up the shell method integral

Since we are revolving the region \(R\) around the y-axis, we can use the shell method formula for the volume: $$V=2\pi\int_{a}^{b}x(y)h(y) dy$$ where \(V\) is the volume, \(a\) and \(b\) are the boundaries of the integration, \(x(y)\) is the distance from the axis of revolution (in this case, the y-axis) to the curve defining the region, and \(h(y)\) is the thickness of the shell (which is equal to the height of the curve). Since we are revolving around the y-axis, we need to express both \(x\) and \(h\) in terms of \(y\). Our original function is given by \(y=x-x^2\), so we can solve for x to find \(x(y)\): $$y=x(1-x)$$ $$x=\frac{y}{1-x}$$ We can now express \(x\) terms of \(y\) by solving for \(x\) in terms of \(y\): $$x = 1 - \sqrt{1-y}$$ Now we need to find the height of the curve \(y=x-x^2\). Since the region is bounded by the x-axis, the height is just the y-coordinate: $$h(y) = y$$ We now have everything needed to set up our integral: $$V=2\pi\int_{0}^{1}(1-\sqrt{1-y})y dy$$

Step 3: Evaluate the integral

Now we just need to evaluate the integral to find the volume of the solid: \begin{align*} V &= 2\pi\int_{0}^{1}(1-\sqrt{1-y})y dy \\ &= 2\pi\left[\int_{0}^{1}y dy - \int_{0}^{1}\sqrt{1-y}y dy\right] \end{align*} The first integral is a simple polynomial integral: $$\int_{0}^{1}y dy = \frac{1}{2}y^2\Bigg|_0^1 = \frac{1}{2}$$ The second integral can be solved using a substitution. Let \(u=1-y\), then \(du=-dy\). When \(y=0\), \(u=1\); when \(y=1\), \(u=0\). The integral becomes: $$\int_{0}^{1}\sqrt{1-y}y dy = -\int_{1}^{0}\sqrt{u}(-du) =\int_{1}^{0}u^{\frac{1}{2}}du$$ $$= \frac{2}{3}u^{\frac{3}{2}}\Bigg|_1^0=-\frac{2}{3}(0-1)=\frac{2}{3}$$ Now, we can combine the results of both integrals: $$V=2\pi\left(\frac{1}{2}-\frac{2}{3}\right)=-\frac{2\pi}{3}$$ Since the volume cannot be negative, we take the absolute value of the result: $$V=\frac{2\pi}{3}$$ The volume of the solid generated when the region \(R\) is revolved about the \(y\)-axis is \(\frac{2\pi}{3}\).