91Ó°ÊÓ

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 \((1,0)\).

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: $$V=2\pi\int_{0}^{1}(1-\sqrt{1-y})y dy$$

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 \(\frac{2\pi}{3}\).