91影视

The curve is$r=4\sin 3胃$

The goal of this task is to determine the area of one curve loop is $r=4\sin 鈦�3胃$

Pointing the curve role="math" $\mathrm{r}=4\sin 鈦�3\mathrm{胃}$

Graph of $r=4\sin 鈦�3胃$

Determine the tangent at the pole.

Substitute r =0, in the curve

$\begin{array}{r}4\sin 鈦�3胃=0\\ \sin 鈦�3胃=0\end{array}$

That is, $3胃=n蟺$

Therefore, $胃=\frac{n蟺}{3}$

From role="math" localid="1651493144503" $\mathrm{n}=0,1,2,3,4\text{and}5\text{.}$

Put n = 0 and 1 in one loop

Tangents at the pole are then calculated role="math" $\mathrm{胃}=0\text{and}\mathrm{胃}=\frac{\mathrm{蟺}}{3}$

The area of the region surrounded by one curve loop can be represented as $\mathrm{A}={鈭�}_0^{\mathrm{蟺}/3}鈥�{鈭�}_0^{\mathrm{r}=4\sin 鈦�3\mathrm{胃}}鈥�\mathrm{谤诲谤诲胃}$

Integrate first with regard to r.

$\mathrm{A}={鈭�}_0^{\mathrm{蟺}/3}鈥�{\left[\frac{{\mathrm{r}}^2}{2}\right]}_0^{4\sin 鈦�3\mathrm{胃}}\mathrm{诲胃}$

Set the boundaries

$\begin{array}{r}\mathrm{A}={鈭�}_0^{\mathrm{蟺}/3}鈥�\left[\frac{(4\sin 鈦�3\mathrm{胃}{)}^2鈭�0}{2}\right]\mathrm{诲胃}\\ \mathrm{A}={鈭�}_0^{\mathrm{蟺}/3}鈥�8{\sin }^2鈦�3\mathrm{胃诲胃}\\ \mathrm{A}={鈭�}_0^{\mathrm{蟺}/3}鈥�4(1鈭�\cos 鈦�6\mathrm{胃})\mathrm{诲胃}\left[2{\sin }^2鈦�3\mathrm{胃}=1鈭�\cos 鈦�6\mathrm{胃}\right]\end{array}$

Integrate in relation to $胃,$

$\mathrm{A}=4{\left[\mathrm{胃}鈭�\frac{1}{6}\sin 鈦�6\mathrm{胃}\right]}_0^{\mathrm{蟺}/3}$

Pointing the limits,

$\begin{array}{r}\mathrm{A}=4\left[\left(\frac{\mathrm{蟺}}{3}鈭�\frac{1}{6}\sin 鈦�2\mathrm{蟺}\right)鈭�0\right]\\ \mathrm{A}=\frac{4}{3}\mathrm{蟺}\end{array}$

As a result, the area of one curve loop$\mathrm{r}=4\sin 鈦�3\mathrm{胃}$is$\mathrm{A}=\frac{4}{3}\mathrm{蟺}$