ardentmed said:
For the first one, I'm completely dumbfounded. I used the SA formula for and ended up getting 2 $π\int_{π/6}^{0} \, cos2x * √1+4sin(2x)^2 ) $
And took u = sin2x, ultimately giving me $ π\int_{√3/2}^{0} \,d (1+ 4u^2)^(1/2) $
I'm lost as to what to do next. I'm almost definite that my u-substitution is incorrect.
Also for the second one, I used the formula for cardiac output, which is mg/ $\int_{b}^{a} \,d $ , which gave me 78.533 by using the Simpson's Rule in the denominator with delta x = 2.
Any ideas?
Thanks in advance.
For the first one:
The formula is the following:
$$S= \int 2 \pi y ds , \ \text{ where } ds=\sqrt{1+(\frac{dy}{dx})^2}dx$$
$\displaystyle{\frac{dy}{dx}=\frac{\cos{(2x)}}{dx}=-2\sin{(2x)} }$
$\displaystyle{\sqrt{1+(\frac{dy}{dx})^2}=\sqrt{1+(-2\sin{(2x)} )^2}=\sqrt{1+4\sin^2{(2x)}}}$
So $$S=\int_0^{\frac{\pi}{6}} 2 \pi y \sqrt{1+4\sin^2{(2x)}}dx=\int_0^{\frac{\pi}{6}} 2 \pi \cos{(2x)} \sqrt{1+4\sin^2{(2x)}}dx$$
Set $u=\sin{(2x)}$, then:
when $x=0 \Rightarrow u=0$
when $x=\frac{\pi}{6} \Rightarrow u=\frac{\sqrt{3}}{2}$
$du=2 \cos{(2x)}dx$
$$S=\int_0^{\frac{\pi}{6}} 2 \pi \cos{(2x)} \sqrt{1+4\sin^2{(2x)}}dx=\int_0^{\frac{\sqrt{3}}{2}} \pi \sqrt{1+4u^2}du=\int_0^{\frac{\sqrt{3}}{2}} \pi \sqrt{1+(2u)^2}du$$
Set $2u=\tan{w}$, then:
when $u=0 \Rightarrow w=0$
when $u=\frac{\sqrt{3}}{2} \Rightarrow \tan{w}=\sqrt{3} \Rightarrow w=\frac{\pi}{3}$
$2du=\frac{1}{\cos^2{w}}dw$
$$S=\int_0^{\frac{\sqrt{3}}{2}} \pi \sqrt{1+(2u)^2}du=\int_0^{\frac{\pi}{3}} \frac{\pi}{2} \sqrt{1+\tan^2{w}}\frac{1}{\cos^2{w}}dw=\int_0^{\frac{\pi}{3}} \frac{\pi}{2} \sqrt{\frac{1}{\cos^2{w}}}\frac{1}{ \cos^2{w}}dw=\int_0^{ \frac{ \pi}{3}} \frac{\pi}{2} \frac{1}{ \cos^3{w}}dw=\frac{\pi}{2} \int_0^{ \frac{ \pi}{3}} \frac{\cos{w}}{ \cos^4{w}}dw=\frac{\pi}{2} \int_0^{ \frac{ \pi}{3}} \frac{\cos{w}}{ (1-\sin^2{w})^2}dw$$
Set $v=\sin{w}$, then:
when $w=0 \Rightarrow v=0$
when $w=\frac{\pi}{3} \Rightarrow v=\frac{\sqrt{3}}{2}$
$dv=\cos{w}dw$
$$S=\frac{\pi}{2} \int_0^{ \frac{ \pi}{3}} \frac{\cos{w}}{ (1-\sin^2{w})^2}dw=\frac{\pi}{2} \int_0^{ \frac{\sqrt{3}}{2}} \frac{1}{ (1-v^2)^2}dv=\frac{\pi}{2} \int_0^{ \frac{\sqrt{3}}{2}} \frac{1}{ (v^2-1)^2}dv= \\ \frac{\pi}{2} \int_0^{ \frac{\sqrt{3}}{2}} (\frac{A}{ (v-1)}+\frac{B}{v+1})^2dv$$
Calculate the values of $A$ and $B$ as followed:
$\displaystyle{\frac{A}{ v-1}+\frac{B}{v+1}=\frac{1}{(v-1)(v+1)}}$
Can you continue from here?
