Using Green's Theorem, calculate the area of the indicated region.
-The circle $\mathbf { r } ( \mathrm { t } ) = ( 11 \cos \mathrm { t } ) \mathbf { i } + ( 11 \sin \mathrm { t } ) \mathbf { j } , 0 \leq \mathrm { t } \leq 2 \pi$
A) $121 \pi$
B) $2 \pi$
C) $11 \pi$
D) $22 \pi$

Question

Quizplus · Accepted Answer

The area of a region bounded by a curve $\mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}$ can be calculated using Green's Theorem by integrating $dA = \frac{1}{2}(xdy - ydx)$. For the given circle $\mathbf{r}(t) = (11\cos t)\mathbf{i} + (11\sin t)\mathbf{j}$, the differential area element becomes $dA = \frac{1}{2}(11\cos t d(11\sin t) - 11\sin t d(11\cos t))$. Integrating this over $t$ from $0$ to $2\pi$ gives the area of the circle as $121\pi$, since the integral simplifies to the area formula $\pi r^2$ with $r = 11$.

Using Green's Theorem, Calculate the Area of the Indicated Region r(t)=(11cos⁡t)i+(11sin⁡t)j,0≤t≤2π\mathbf { r } ( \mathrm { t } ) = ( 11 \cos \mathrm { t } ) \mathbf { i } + ( 11 \sin \mathrm { t } ) \mathbf { j } , 0 \leq \mathrm { t } \leq 2 \pir(t)=(11cost)i+(11sint)j,0≤t≤2π

Using Green's Theorem, Calculate the Area of the Indicated Region $\mathbf { r } ( \mathrm { t } ) = ( 11 \cos \mathrm { t } ) \mathbf { i } + ( 11 \sin \mathrm { t } ) \mathbf { j } , 0 \leq \mathrm { t } \leq 2 \pi$