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Find the volume generated by revolving the area bounded by x2 - 2x + y = 3, the y-axis and the line y = 4 about x = 0. |
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Answer» \(\because\) Revolution is about y - axis (or x = 0 line) Volume = \(\int\limits^{y_1}_{y_2}\pi x^2\,dy\) Given curve is x2 - 2x + y = 3 ⇒ y = 3 - x2 + 2x = 4 - (x2 - 2x + 1) = 4 - (x - 1)2 At x = 0 on y-axis, y = 4 - 1 = 3 \(\therefore\) y1 = 3, y2 = 4 \(\therefore\) Generated volume = \(\int\limits^4_3\pi x^2\,dy\) \(=\int\limits^4_3\pi (-\sqrt{4-y}+1)\,dy\) \(=\, ^\pi \left(\frac23(4-y)^{\frac32}+y\right)^4\) \(= \pi \left(\frac23 \times0+4-\frac23 \times 1 -3\right)\) \(= \pi \left(1 - \frac23\right)\) \(= \frac {\pi}3\) cubic units. |
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