Question

Find the area of the region enclosed by one loop of the curve.

\({\rm{r = 4cos3\theta }}\)

Short answer

The loop's enclosing region\(\frac{{{\rm{4\pi }}}}{{\rm{3}}}.\)\(\frac{{{\rm{4\pi }}}}{{\rm{3}}}.\)

Step-by-step solution

One of the loops will simply enclose an area.

\(\begin{aligned}{*{20}{r}}{{\rm{r = 4cos(3\theta )}}}\\{_{\rm{a}}^{\rm{b}}\frac{{{{\rm{r}}^{\rm{2}}}}}{{\rm{2}}}{\rm{d\theta }}}\end{aligned}\)

Calculate the integration limits \(a,b\)and then use the integral to find the area contained by one of the loops.\(a,b\)

To obtain the integration limits.

Locate two successive \({\rm{\theta }}\)values for which \({\rm{r}}\)is zero.

Graph \({\rm{y = 4cos3x}}\)as depicted below

\({\rm{ - }}\frac{{\rm{\pi }}}{{\rm{6}}}\)And\(\frac{{\rm{\pi }}}{{\rm{6}}}\)for which there are two consecutive values

\({\rm{4cos3x}}\)Is nil As a result, may integrate from\({\rm{ - }}\frac{{\rm{\pi }}}{{\rm{6}}}\)to\(\frac{{\rm{\pi }}}{{\rm{6}}}\).

You can also incorporate information from\(\frac{{\rm{\pi }}}{{\rm{6}}}\)to\(\frac{{\rm{\pi }}}{{\rm{2}}}\)

The only requirement is that you integrate between two consecutives\({\rm{0}}\) values with \({\rm{r = 0}}\).

The sole stipulation is that you must integrate between two numbers that are consecutive.  

Step 4:  One of the loops encloses a certain area.

\(\begin{aligned}{c}{\rm{\pi /6}}{{\rm{r}}^{\rm{2}}}\frac{{{{\rm{r}}^{\rm{2}}}}}{{\rm{2}}}{\rm{d\theta }}\\{\rm{ - \pi /6}}\frac{{{{{\rm{(4cos(3\theta ))}}}^{\rm{2}}}}}{{\rm{2}}}{\rm{d\theta }}\\{\rm{ - \pi /6\pi /6}}\\{\rm{ - \pi /6}}\\{\rm{\pi /6}}\;\;\;\frac{{{\rm{1 + co}}{{\rm{s}}^{\rm{2}}}{\rm{(3\theta )d\theta }}}}{{{\rm{cos(6\theta )}}}}\\{\rm{ - \pi /6}}\end{aligned}\)

The loop's enclosed area is\(\frac{{{\rm{4\pi }}}}{{\rm{3}}}.\)