Consider the parabola given by the equation y^2-2x-6y+11=0
How to get it in Standard Form and find the vertex and focus

Find the center of the ellipse
5x^2+2y^2-20x+24y+82=0

Find the length of the curve over the given interval
r=8(1+cos(angle), 0<angle<2(3.124)

Find the area of the region.
interior of r=5(1+sin(angle))

and

One petal of r=cos(5(angle))

Find the area of the region.
interior of r=5(1+sin(t))

and

One petal of r=cos(5t)

Is this one problem or two? I am going to treat it as two different regions.

\frac{25}{2}\int_{0}^{2\pi}(1+sin(t))^{2}dt

=\frac{25}{2}\int_{0}^{2\pi}\left[2sin(t)dt-\frac{1}{2}cos(2t)+\frac{3}{2}\right]dt

Now, integrate away.

As for the rose, to find the limits of integration, solve cos(5t)=0

We find t=\frac{\pi}{10}

Multiply by 2 because of symmetry.

\int_{0}^{\frac{\pi}{10}}(cos(5t))^{2}dt=\int_{0}^ {\frac{\pi}{10}}\left[\frac{1+cos(10t)}{2}\right]

Find the length of the curve over the given interval
r=8(1+cos(t)), \;\ 0<t<2{\pi}



r=8(1+cos(t))

Polar arc length is given by:

\int_{\alpha}^{\beta}\sqrt{r^{2}+\left(\frac{dr}{d t}\right)^{2}}dt

\frac{dr}{dt}=-8sin(t)

I assume you are integrating from 0 to 2{\pi}

\int_{0}^{2\pi}\sqrt{(8(1+cos(t)))^{2}+(-8sin(t))^{2}}dt

\int_{0}^{2\pi}\sqrt{128(cos(t)+1)}dt

This integral can be a little tricky. As a matter of fact, most calculators grunt trying to do it, but it ain't that bad if we use some tricks.

8\sqrt{2}\int_{0}^{2\pi}\sqrt{1+cos(t)}dt

\sqrt{\frac{1+cos(t)}{1}\cdot\frac{1-cos(t)}{1-cos(t)}}

=\sqrt{\frac{1-cos^{2}(t)}{1-cos(t)}}

=\sqrt{\frac{sin^{2}(t)}{1-cos(t)}}

=\frac{sin(t)}{\sqrt{1-cos(t)}}

Now, let u=1-cos(t), \;\ du=sin(t)dt

We end up with something easy:

\int\frac{1}{\sqrt{u}}du=2\sqrt{u}

Resub:

2\sqrt{1-cos(t)}

Let's integrate from 0 to Pi and multiply by 2 because using 2Pi presents a conundrum. Do you see why?

2\cdot 8\sqrt{2}\sqrt{1-cos(\pi)}=64

There it is.

Find the center of the ellipse
5x^{2}+2y^{2}-20x+24y+82=0




Complete the square.

5x^{2}-20x+2y^{2}+24y=-82

5(x^{2}-4x+4)+2(y^{2}+12y+36)=-82+20+72

5(x-2)^{2}+2(y+6)^{2}=10

Now, can you finish and get into the form of an ellipse? Just one more step.

Though, you can easily see the center coordinates as it is now.

Consider the parabola given by the equation y^{2}-2x-6y+11=0
How to get it in Standard Form and find the vertex and focus



As with the ellipse, complete the square.

The standard form of this parabola is x=a(y-h)^{2}+k, where (h,k) are the vertex coordinates.

It is a parabola opening to the right symmetric about y=3. The directrix is at x=1/2.

y^{2}-6y-2x=-11

(y^{2}-6y+9)-2x=-11+9

(y-3)^{2}+2=2x

x=\frac{1}{2}(y-3)^{2}+1

The focus can be found by using p=\frac{1}{4a}.