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Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

$y = \sin x$, $y = \cos x$, $0 < x < \frac{\pi}{4}$; rotated about the line $y = -1$

So...

outer radius: $1 + \cos x$

inner radius: $1 + \sin x$

So here's my setup:

$$\pi \int_0^\frac{\pi}{4} (1 + \cos x)^2 - (1 + \sin x)^2 \, dx$$

$$\pi \int_0^\frac{\pi}{4} 1 + 2 \cos x + \cos^2x - ( 1 + 2 \sin x + \sin^2 x ) \, dx$$

Is this setup right?

But then I'm stuck here on the next step:

$$\pi \int_0^{\frac{\pi}{4}} (2 \cos x - 2 \sin x + \cos^2 x + \sin^2 x \,) dx$$

From here, I need to find the antiderivatives. but what are the antiderivatives of $\cos^2 x$? Also, how do I draw this graph using mathjax?

Michael Hardy
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Jwan622
  • 5,704
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    The integrand in the last step should be $2\cos x-2\sin x+\cos^2x-\sin^2x$, which is equal to $2\cos x-2\sin x+\cos2x$ – CY Aries Apr 17 '18 at 15:43
  • You have $\cos^2x+\sin^2x,$ which you should know is equal to $1,$ but you ought to have $\cos^2x-\sin^2x,$ and you should recall that that is $\cos(2x). \qquad$ – Michael Hardy Apr 17 '18 at 17:15

1 Answers1

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Hint

As pointed out in the comments you have: $$\cos^2(x)-\sin^2(x)=\cos(2x)$$ In all generality when you have $\cos^2$ or $\sin^2$ you can use the formulas: $$\cos^2(x)=\frac{1+\cos(2x)}{2}$$ $$\sin^2(x)=\frac{1-\cos(2x)}{2}$$ from where it is easy to compute anti derivatives.

For the graph I don't think there is a easy way to fraw graph using mathjax but in LaTeX you can use pgfplots:

\documentclass{standalone}
\usepackage{pgfplots}


\usepgfplotslibrary{fillbetween}
\usetikzlibrary{patterns}

\begin{document}

\begin{tikzpicture}
\begin{axis}[axis lines=middle,
            xlabel=$x$,
            ylabel=$y$,
            enlargelimits,
            ytick=\empty,
            xtick={0.785},
            xticklabels={$\pi/4$}]
\addplot[name path=S,blue,domain={0:pi/4}] {sin(deg(x))} node[pos=0.5, below]{$y=\sin(x)$};

\addplot[name path=C,red,domain={0:pi/4}] {cos(deg(x))}node[pos=0.5, above]{$y=\cos(x)$};

\addplot[pattern=north west lines, pattern color=black!20]fill between[of=S and C]
;


\end{axis}
\end{tikzpicture}
\end{document}

result

Delta-u
  • 6,038
  • I don't see the antiderivatives clearly. What are they and why? They don't seem to fall under my chart of antiderivative rules. – Jwan622 Apr 17 '18 at 16:35
  • For $\cos^2(x)$ you have: $$\frac{1+\cos(2x)}{2}=\frac{1}{2}+\frac{1}{4} 2 \cos(2x)$$ as the anti derivative of $1$ is $x+C$ and of $2 \cos(2x)$ is $\sin(2x)$ you can compute the result: $$\frac{x}{2}+\frac{1}{4}\sin(2x)+C$$ It is the same for $\sin^2$.

    In fact for your case you only have to compute the antiderivative of $\cos(2x)$ whici is $\frac{1}{2} \sin(2x)$.

     – Delta-u Apr 17 '18 at 16:40
  • Is there a rule for this? Or did you kind of just intuit this? – Jwan622 Apr 17 '18 at 18:48
  • Yes there is rules: use the linearity of the antiderivative and if $F(x) $ is an antiderivative of $f(x) $ then an antiderivative of $f (ax+b) $ is $1/a F (ax +b) $ ( if $a\neq 0$). – Delta-u Apr 17 '18 at 19:27