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試求\(y=-x^2-3x+6\)和\(x+y-3=0\)所圍成的區域繞\(x=2\)所得的旋轉體體積為
[解答]
可以用Pappus 定理
\( \displaystyle -x^2-3x+6 = -x+3 \; \Rightarrow \; x = -3,1 \quad , \quad A = \int_{-3}^{1} [(-x^2-3x+6) - (-x+3)] dx = \frac{32}{3} \)
\( \displaystyle \overline{X} = \frac{ \displaystyle \int \int x dA }{ \displaystyle \int \int dA } = \frac{1}{A} \int_{-3}^{1} (-x^3 - 2x^2 + 3x) dx = \frac{ \displaystyle -\frac{32}{3} }{ \displaystyle \frac{32}{3} } = -1 \)
\( \displaystyle V = A \times 2\pi R = \frac{32}{3} \times 2 \pi \times [2 - (-1)] = 64\pi \)
