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

The given curve is rotated about the -axis. Find the area of the resulting surface. $$\displaystyle{y}={\frac{{{1}}}{{{4}}}}{x}^{{{2}}}-{\frac{{{1}}}{{{2}}}}{\ln{{x}}},{1}\le{x}\le{2}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Terry Ray
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
###### Not exactly what you’re looking for?
Corgnatiui
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 $$\displaystyle={\frac{{{10}\pi}}{{{3}}}}$$
star233

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