1. ## Area Volume

1. Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

y= sqrt(30-x) and y=-x i inserted in graphing calculator but no graph???

2. Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

I graphed and is from 0 to 3.

y= 4x^3/3 and y=1/3x(x^2+9)

i got 54. did i do it right?

2. I would integrate with respect to y, becaus if you did it with respect to x

x=+/- sqrot(30-y^2) is not a function but a relation.

Am I right?

3. For number 2

4x^2=X^2+3

3x^2=3

x^2=1 so x=1,-1

so the interval to look at is [-1,1] Right?

4. You don't need a graphing calculator for this. It's best to have your equations for the first problem as $x=f(y)$. So we have $x=30-y^{2}$ and $x=-y$ Set the two equal to each other to find the points of intersection:

$30-y^{2}=-y$
$0=y^{2}-y-30$
$0=(y-6)(y+5)$
$y=6,y=-5$

You should be able to sketch of graph of this and know that $30-y^{2} \ge -y$ on the interval of $[-5,6]$, so your integral should be:

$\int_{-5}^{6}30-y^{2}+y\,\,dy$

For the second problem, it's obvious we should have our functions as $y=f(x)$. Again, set the two equations equal to each other:

$4x^{2}=x^{2}+3$
$3x^{2}=3$
$x=-1,x=1$

Again, sketching the graphs of these are easy, and you can tell $x^{2}+3 \ge 4x^{2}$ on $[-1,1]$, so your integral would be:

$\int_{-1}^{1}3-3x^{2}\,\,dx$

5. Thanks for helping guys. I really appreciated.