Hi, there!
Due to my fickle nature, I’ve never moved on to reading the second volume of any literature masterpieces and English magazines. You know, I always get bored when I have almost finished reading the first one.
I guess the author will also get in an old rut and get out of freshness and insightfulness in the second volume. My bottom line is that the author or publisher might want to avoid thinking the business model is lucrative.
Once I have read a Japanese voluminous or prolific writer’s mystery novels, but most of the series end up with corny endings, and I cannot bask in the afterglow of so-called “magnum opuses.”
The heyday of magnificent greatest works is apparently but reluctantly a thing of the past.
Today, I try to find the volume of the solid obtained by rotating the region bounded by a quadratic function and the two specific points at the x-axis; Don’t worry; it’s not voluminous.
Determine the volume of the solid obtained by rotating the region bounded by \(y = x^{2} – 2x + 3\), \(x = 2\), \(x = 5\), and the \(x\)-axis about the \(x\)-axis.
Here is my solution to the problem.
\(A(x) = π(x^{2} – 2x + 3)^{2} = π(x^{4} -4x^{3} + 10x^{2} – 12x + 9)\)
\(V = \int_{a}^{b}A(x)dx\)
\(= π\int_{2}^{5}x^{4} – 4x^{3} + 10x^{2} – 12x + 9\,dx\)
\(＝π[\frac{1}{5}x^{5} – x^{4} + \frac{10}{3}x^{3} – 6x^{2} + 9x]_{2}^{5}\)
\(= \frac{889}{3}π\)
I’m unsure whether my answer is correct because I reached the answer so quickly as if I enjoy “Nagashi Soumen” or fine white noodles that flow in a small flume, seasoned in summer in Japan. I would appreciate it very much if you could report to me if my calculations were wrong. Thanks.
Stay tuned, and expect to see my next post.
Keep well.
Frank Yoshida
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【グローバルビジネスで役立つ数学】でもっと学習する b＾＾)
【参考図書】『もう一度高校数学』（著者：高橋一雄氏）株式会社日本実業出版社
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