How to find the surface area of rotary solid.

Question

What is the surface area of solid generated by rotating the parametric curve $x=\cos\left(\frac{\pi u}{2}\right)\ $ ($\forall \ \ 0\le u\le 1$) about x-axis through $360^\circ$ ?

My try:

I treated $u$ as an independent variable $x$ & variable $x$ as dependent variable $y$ then the equation can be re-written as

$y=\cos\left(\frac{\pi x}{2}\right)$

$y'=-\frac\pi2 \sin\left(\frac{\pi x}{2}\right)$

The surface area of rotary solid

$=\int_0^1 2\pi y\sqrt{1+(y')^2}\ dx$

$=\int_0^1 2\pi \cos\left(\frac{\pi x}{2}\right)\sqrt{1+\left(-\frac\pi2 \sin\left(\frac{\pi x}{2}\right)\right)^2}\ dx$

$=\frac8\pi\int_0^1 \sqrt{1+\left(\frac\pi2 \sin\left(\frac{\pi x}{2}\right)\right)^2}\ d\left(\frac\pi2\sin\left(\frac{\pi x}{2}\right)\right)$

$=\frac12\left(\sqrt2+\ln(1+\sqrt2)\right)$

But my teacher says that the correct answer is $2\pi$ sq.unit, I don't where I am wrong. Please tell me where I am wrong & help me solve it. Thanks.

Answer 1

The data you have given seems not complete.

Assuming as parametric curve

• $x=\cos\left(\frac{\pi u}{2}\right)$
• $y=u$

we have

$$x=\cos\left(\frac{\pi y}{2}\right)\implies y=\frac 2 \pi\arccos x \implies y'=-\frac 2 \pi\frac1{\sqrt{1-x^2}}$$

and therefore we obtain (see numerical evaluation here)

$$S=\int_0^1 2\pi \cdot \frac 2 \pi\arccos x\sqrt{1+\frac4{\pi^2}\frac1{1-x^2}}\,dx\approx 5.14$$

Assuming, according to your comment, as parametric curve

• $x=\cos\left(\frac{\pi u}{2}\right)$
• $y=\sin\left(\frac{\pi u}{2}\right)$

we have that the cordinates describe a quarter of a circle and in that case we obtain

$$V=\frac12 \cdot 4\pi=2\pi$$

Answer 2

first plot x vs u graph by varying u from 0 to 1 (you will get a quardant with radius 1 )

then, rotate your curve about x axis . you'll get hemisphere of radius $1$ giving surface area =$2\pi $ sq. units

Answer 3

This is not a parametric curve. When you change your variables from $u$ to $x$ and $x$ to $y$, then you also have to change them in the phrase "about the $x$-axis" to "about the $y$-axis. So your integral should be

$$\int_0^1 2\pi x \sqrt{1+(y')^2} \; dx.$$

And that has a numerical value of about $5.1$, so there's no way the right answer is $2\pi.$