Question

Sketch the curve and find the area that it encloses.

\({\rm{r = 2sin\theta }}\)

Short answer

a complete remedy\({\rm{\pi }}{\rm{.}}\)

Step-by-step solution

Multiply both sides of the graph by \({\rm{r}}{\rm{.}}\)

\({\rm{r = 2sin\theta }}\)

\({{\rm{r}}^{\rm{2}}}{\rm{ = 2rsin\theta }}\)

Multiply both sides by \({\rm{r}}\)in order to create a graph.

\({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = 2y}}\)

As an alternative to

\({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ - 2y = 0}}\)

Calculate the value of \({\rm{0}}\)

Finish the square by adding \({\rm{1}}\) to both sides.

\({{\rm{x}}^{\rm{2}}}{\rm{ + (y - 1}}{{\rm{)}}^{\rm{2}}}{\rm{ = 1}}\)

Add \({\rm{1}}\)to each sides of the square to complete it.

\(\begin{aligned}{c}{{\rm{y}}^{\rm{2}}}{\rm{ - 2y + 1 = 1}}\\{{\rm{(y - 1)}}^{\rm{2}}}{\rm{ = 1}}\end{aligned}\)

\(\frac{{\rm{1}}}{{\rm{2}}}_{\rm{a}}^{\rm{b}}{{\rm{r}}^{\rm{2}}}{\rm{d\theta }}\;\;\;\)

Circle graph with radius \({\rm{1}}\) and centred at \(\left( {{\rm{0,1}}} \right){\rm{.}}\)

Multiply the number of squares by the number of squares.

To get the area, multiply the number of squares by the number of squares in the to obtain\({\rm{r}}\)

\(\frac{{\rm{1}}}{{\rm{2}}}_{\rm{0}}^{\rm{\pi }}{\rm{4si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta d\theta }}\)

Because the graph is in quadrants I and II, use the area formula with bounds\({\rm{0}}\)to\({\rm{\pi }}\).

\({\rm{2}}\int_{\rm{0}}^{\rm{\pi }} {\frac{{{\rm{1 - cos2\theta }}}}{{\rm{2}}}} {\rm{d\theta }}\)

Move\({\rm{4}}\)away from the integral.

Then make a substitution.

\({\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}\)

Formula for a half-angle.

\(\frac{{{\rm{1 - cos2\theta }}}}{{\rm{2}}}\)

\(\left( {{\rm{\theta - }}\frac{{{\rm{sin2\theta }}}}{{\rm{2}}}} \right)_{\rm{0}}^{\rm{\pi }}\)

\(\frac{{\rm{1}}}{{\rm{2}}}\) Expand your horizons

Then, with regard to\({\rm{\theta ,}}\)integrate.

\({\rm{(\pi - 0) - (0 - 0) = \pi }}\)

A comprehensive solution \({\rm{\pi }}\).