The given curve is rotated about the -axis. Find the

veksetz 2022-01-12 Answered
The given curve is rotated about the -axis. Find the area of the resulting surface. \(\displaystyle{y}={\frac{{{1}}}{{{4}}}}{x}^{{{2}}}-{\frac{{{1}}}{{{2}}}}{\ln{{x}}},{1}\le{x}\le{2}\)

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Expert Answer

Terry Ray
Answered 2022-01-13 Author has 5676 answers
Step 1
\(\displaystyle{S}_{{{y}}}={\int_{{{a}}}^{{{b}}}}{2}\pi{x}\sqrt{{{\left({\left.{d}{x}\right.}\right)}^{{{2}}}+{\left({\left.{d}{y}\right.}\right)}^{{{2}}}}}\)
We will use the following version of the formula
\(\displaystyle{S}_{{{y}}}={\int_{{{a}}}^{{{b}}}}{2}\pi{x}\sqrt{{{1}+{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\right)}^{{{2}}}}}{\left.{d}{x}\right.}\)
The equation of the curve is \(\displaystyle{y}={\frac{{{1}}}{{{4}}}}{x}^{{{2}}}-{\frac{{{1}}}{{{2}}}}{\ln{{x}}}\)
Differentiate both sides with respect to x
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{1}}}{{{4}}}}\cdot{2}{x}^{{{2}-{1}}}-{\frac{{{1}}}{{{2}}}}\cdot{\frac{{{1}}}{{{x}}}}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}}}{{{2}}}}-{\frac{{{1}}}{{{2}{x}}}}\)
Substitute this in the formula for \(\displaystyle{S}_{{{y}}}\), to get
\(\displaystyle{S}_{{{y}}}={\int_{{{1}}}^{{{2}}}}{2}\pi{x}\sqrt{{{1}+{\left[{\frac{{{x}}}{{{2}}}}-{\frac{{{1}}}{{{2}{x}}}}\right]}^{{{2}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle{S}_{{{y}}}={\int_{{{1}}}^{{{2}}}}{2}\pi{x}\sqrt{{{1}+{\left[{\frac{{{x}}}{{{2}}}}\right]}^{{{2}}}+{\left[{\frac{{{1}}}{{{2}{x}}}}\right]}^{{{2}}}-{2}{\left[{\frac{{{x}}}{{{2}}}}\right]}{\left[{\frac{{{1}}}{{{2}{x}}}}\right]}}}{\left.{d}{x}\right.}\)
Pay attention to the red term
\(\displaystyle{S}_{{{y}}}={\int_{{{1}}}^{{{2}}}}{2}\pi{x}\sqrt{{{1}+{\left[{\frac{{{x}}}{{{2}}}}\right]}^{{{2}}}+{\left[{\frac{{{1}}}{{{2}{x}}}}\right]}^{{{2}}}-{\frac{{{1}}}{{{2}}}}}}{\left.{d}{x}\right.}\)
On simplification, the red term reduces to \(\displaystyle-{\frac{{{1}}}{{{2}}}}\)
Now, we will combine this with 1 that is already present inside the square root, so that \(\displaystyle-{\frac{{{1}}}{{{2}}}}\) now becomes \(\displaystyle{\frac{{{1}}}{{{2}}}}\).
Step 2
\(\displaystyle{S}_{{{y}}}={\int_{{{1}}}^{{{2}}}}{2}\pi{x}\sqrt{{{\left[{\frac{{{x}}}{{{2}}}}\right]}^{{{2}}}+{\left[{\frac{{{1}}}{{{2}{x}}}}\right]}^{{{2}}}+{\frac{{{1}}}{{{2}}}}}}{\left.{d}{x}\right.}\)
But, we know that \(\displaystyle{\frac{{{1}}}{{{2}}}}\) can be rewritten as \(\displaystyle{2}{\left[{\frac{{{x}}}{{{2}}}}\right]}{\left[{\frac{{{1}}}{{{2}{x}}}}\right]}\)
\(\displaystyle{S}_{{{y}}}={\int_{{{1}}}^{{{2}}}}{2}\pi{x}\sqrt{{{\left[{\frac{{{x}}}{{{2}}}}\right]}^{{{2}}}+{\left[{\frac{{{1}}}{{{2}{x}}}}\right]}^{{{2}}}+{2}{\left[{\frac{{{x}}}{{{2}}}}\right]}{\left[{\frac{{{1}}}{{{2}{x}}}}\right]}}}{\left.{d}{x}\right.}\)
Notice that the expression inside the square root is now in the perfect square form
\(\displaystyle{S}_{{{y}}}={\int_{{{1}}}^{{{2}}}}{2}\pi{x}\sqrt{{{\left[{\frac{{{x}}}{{{2}}}}+{\frac{{{1}}}{{{2}{x}}}}\right]}^{{{2}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle{S}_{{{y}}}={\int_{{{1}}}^{{{2}}}}{2}\pi{x}{\left[{\frac{{{x}}}{{{2}}}}+{\frac{{{1}}}{{{2}{x}}}}\right]}{\left.{d}{x}\right.}\)
\(\displaystyle{S}_{{{y}}}={\int_{{{1}}}^{{{2}}}}\pi{\left[{x}^{{{2}}}+{1}\right]}{\left.{d}{x}\right.}\)
Use the power rule for integration
\(\displaystyle{S}_{{{y}}}=\pi{{\left[{\frac{{{x}^{{{2}+{1}}}}}{{{2}+{1}}}}+{x}\right]}_{{{1}}}^{{{2}}}}\)
\(\displaystyle{S}_{{{y}}}=\pi{\left[{\frac{{{2}^{{{3}}}}}{{{3}}}}+{2}\right]}-\pi{\left[{\frac{{{1}^{{{3}}}}}{{{3}}}}+{1}\right]}\)
\(\displaystyle{S}_{{{y}}}=\pi{\left[{\frac{{{7}}}{{{3}}}}+{1}\right]}={\frac{{{10}\pi}}{{{3}}}}\)
Answer: \(\displaystyle{S}_{{{y}}}={\frac{{{10}\pi}}{{{3}}}}\) square units
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Corgnatiui
Answered 2022-01-14 Author has 892 answers
Step 1
\(\displaystyle{y}={\frac{{{1}}}{{{4}}}}{x}^{{{2}}}-{\frac{{{1}}}{{{2}}}}{\ln{{x}}}\) (1)
where f is positive and has a continuous derivative, we define the surface area of the surface obtained by rotating the curve \(\displaystyle{y}={\left({x}\right)},{a}\le{x}\le{b}\), about the y-axis as
\(\displaystyle{s}={\int_{{{a}}}^{{{b}}}}{2}\pi{x}\sqrt{{{1}+{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\right)}^{{{2}}}}}{\left.{d}{x}\right.}\)
Step 2
differentiate (1) with respect to x
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{1}}}{{{2}}}}{x}-{\frac{{{1}}}{{{2}{x}}}}\)
\(\displaystyle{\left({\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}\right)}^{{{2}}}={\left({\frac{{{1}}}{{{2}}}}{x}-{\frac{{{1}}}{{{2}{x}}}}\right)}^{{{2}}}\)
\(\displaystyle{\left({\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}\right)}^{{{2}}}={\left({\frac{{{1}}}{{{4}}}}{x}^{{{2}}}-{\frac{{{1}}}{{{2}}}}+{\frac{{{1}}}{{{4}{x}^{{{2}}}}}}\right)}\)
\(\displaystyle{s}={\int_{{{1}}}^{{{2}}}}{2}\pi{x}\sqrt{{{1}+{\frac{{{1}}}{{{4}}}}{x}^{{{2}}}-{\frac{{{1}}}{{{2}}}}+{\frac{{{1}}}{{{4}{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle{s}={\int_{{{1}}}^{{{2}}}}{2}\pi{x}\sqrt{{{\frac{{{1}}}{{{4}}}}{x}^{{{2}}}+{\frac{{{1}}}{{{2}}}}+{\frac{{{1}}}{{{4}{x}^{{{2}}}}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{{1}}}^{{{2}}}}{2}\pi{x}\sqrt{{{\left({\frac{{{1}}}{{{2}}}}{x}+{\frac{{{1}}}{{{2}{x}}}}\right)}^{{{2}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{{1}}}^{{{2}}}}{2}\pi{x}{\left({\frac{{{1}}}{{{2}}}}{x}+{\frac{{{1}}}{{{2}{x}}}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle=\pi{\int_{{{1}}}^{{{2}}}}{2}{\left({x}^{{{2}}}+{1}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle=\pi{{\left[{\frac{{{1}}}{{{3}}}}{x}^{{{2}}}+{x}\right]}_{{{1}}}^{{{2}}}}\)
\(\displaystyle=\pi{\left({\frac{{{8}}}{{{3}}}}+{2}-{\frac{{{1}}}{{{3}}}}-{1}\right)}\)
The area of the surface \(\displaystyle={\frac{{{10}\pi}}{{{3}}}}\)
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star233
Answered 2022-01-15 Author has 0 answers

\(y'=\frac{1}{2}x-\frac{1}{2x} \\A=2 \pi \int_{1}^{2} \times \sqrt{1+(\frac{1}{2}x-\frac{1}{2x})^{2}} dx \\=2 \pi \int_{1}^{2} \times \sqrt{1+\frac{1}{4x^{2}}-\frac{1}{2}+\frac{1}{4x^{2}}} dx \\=2 \pi \int_{1}^{2} \times \sqrt{\frac{1}{2}+\frac{1}{4x^{2}}+\frac{1}{4x^{2}}} dx \\=2 \pi \int_{1}^{2} \sqrt{\frac{1}{2}x^{2}+\frac{1}{4}x^{4}+\frac{1}{4x^{2}}} dx \\=2 \pi \int_{1}^{2} \sqrt{(\frac{1}{2}x^{2}+\frac{1}{2})^{2}} dx \\=2 \pi \int_{1}^{2} \frac{1}{2}x^{2}+\frac{1}{2} dx \\=2 \pi [\frac{1}{6}x^{3}+\frac{1}{2}x]_{1}^{2} \\=2 \pi [\frac{1}{6}(2)^{3}+\frac{2}{2}-\frac{1}{6}-\frac{1}{2}]_{1}^{2} \\=2 \pi (\frac{4}{3}+\frac{3}{3}-\frac{1}{6}-\frac{3}{6}) \\=2 \pi (\frac{1}{3}-\frac{4}{6}) \\=2 \pi (\frac{42}{18}-\frac{12}{18}) \\=2 \pi (\frac{30}{18})=\frac{10}{3} \pi\)

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