Question

Calculus 3

Answer 1

Evaluate exactly \[\int\limits_{}^{}\int\limits_{R}^{} x \sqrt{{x^2+y^2}} dA\]for the region R in the plane bounded
above by line y=-x and below by the circle x^2+ y^2-2 y=0

Answer 2

Anybody please?

IS that an equation of a circle? x^2+ y^2-2 y=0?

Answer 3

yup. But you're going to have to complete the square to make sense of it.

Answer 4

But I have Y and x in the equation.

Answer 5

recall that the equation of a circle is written in the form (x-h)^2+(y-k)^2=r^2
upon expansion we get
x^2-2xh+h^2+y^2-2ky+k^2=r^2
find h and k, along with r... and then rewrite the equation back in the standard form of the circle

Answer 6

I strongly suggest that you sketch the area in the xy plane over which you're to integrate. You do need to find the equation of the circle in question, so as to find its center and radius. Graph that circle. Then draw the line y=-x on the same set of axes.

Answer 7

Yes, x^2+ y^2-2 y=0 is the equation of a circle.
Using "completing the square," you must re-write it in the form \[(x-h)^2+(y-k)^2=r^2\]

Answer 8

The point (h,k) is the center of the circle, and r is the radius.

You OK with "completing the square?"

Answer 9

not with that equation with x and y

Answer 10

*

Answer 11

worry about a single variable.

Answer 12

besides, x^2 is already completed. h=0 in this instance; you should be able to recognize that.

Answer 13

Note that x^2 stands by itself in the given equation x^2+ y^2-2 y=0. As inkyvoyd states, this means that we already know the x-coordinate of the center of the circle: it is 0.

Note that x^2=(x-0)^2.
Now, tackle that y^2-2y. To complete the square, add 1, and then subtract 1, to y^2-2y. Would you do that, please? Show your result. Re-write the original equation x^2+ y^2-2 y=0.

Answer 14

is it \[(y-1)^2-1\]

Answer 15

ya now show the entire equation and rewrite it so you get (x-h)^2+(y-k)^2=r^2

Answer 16

what is k and r

Answer 17

well, from what you've already done, k=1.

Answer 18

you should put what you've done to complete the square in your original equation...

Answer 19

then move the constant to the other side to determine the radius squared, or r

Answer 20

Dont get. put what in the original equation?

Answer 21

so you started with x^2+ y^2-2 y=0 right?
what do you get after you complete the square?

Answer 22

Theoretical answer: (h,k) represents the center of the circle in the xy plane, and r represents the radius of this circle.

Practical answer: You want k and r values.

To obtain them, you MUST copmpelte the square in

x^2+ y^2-2 y=0.

Have you performed "completing the square" before?

Need to see your work.

Start with x^2+ y^2-2 y=0,

ignore the x^2 for now, and concentrate on y^2-2y. How would you "complete the square" without changing the value of x^2+ y^2-2 y=0 ?

Answer 23

r=1

Answer 24

@khally92 , we are looking for the entire equation...

Answer 25

It's true that r=1, but inkyvoyd and I are asking you to show clearlyl how and where you got r=1. Please share your work with us .

Answer 26

\[y^2-2y+0,a=1 b=-2 and c=0 a(y+d)^2+e=0 (y-1)^2 -1\]



put into equation \[x^2+(y-1)^2-1=0 \]

If you take 0 to the other side gives you r=1 h=0 and k=1?

Answer 27

I mean 1 not 0

Answer 28

Looks good. So what does this tell us about our circle? What is the center, and what is the radius?