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Exercise 9.9 · Q4

Q.The region enclosed between the graphs of y=xy=x and y=x2y=x^2 is denoted by RR. Find the volume generated when RR is rotated through 360∘360^\circ about xx-axis.

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Since RR lies between two curves rather than between a single curve and the axis, revolving it about the xx-axis produces a washer (annular) cross-section, not a solid disc; the volume is the outer-disc volume minus the inner-disc volume.

Step 1. Find where y=xy=x and y=x2y=x^2 intersect. Setting x=x2x=x^2 gives x2−x=0⇒x(x−1)=0⇒x=0x^2-x=0\Rightarrow x(x-1)=0\Rightarrow x=0 or x=1x=1. So RR is bounded between x=0x=0 and x=1x=1.

Step 2. Determine which curve is farther from the xx-axis on (0,1)(0,1). For 0<x<10<x<1, x>x2x>x^2 (e.g. at x=12x=\tfrac12: 12>14\tfrac12>\tfrac14), so the line y=xy=x lies above the parabola y=x2y=x^2 throughout RR. On revolution about the xx-axis, y=xy=x therefore sweeps the outer radius and y=x2y=x^2 the inner radius.

Step 3. Set up the washer volume. For a washer, V=π∫ab[Router2−Rinner2]dx=π∫01[x2−(x2)2]dx=π∫01(x2−x4) dxV=\pi\displaystyle\int_a^b\left[R_{\text{outer}}^2-R_{\text{inner}}^2\right]dx=\pi\int_0^1\left[x^2-(x^2)^2\right]dx=\pi\int_0^1(x^2-x^4)\,dx.

Step 4. Integrate. ∫01(x2−x4) dx=[x33−x55]01=13−15=5−315=215\displaystyle\int_0^1(x^2-x^4)\,dx=\left[\dfrac{x^3}{3}-\dfrac{x^5}{5}\right]_0^1=\dfrac13-\dfrac15=\dfrac{5-3}{15}=\dfrac{2}{15}. …

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