Find the area of the region bounded by x=y^2 and y=x-2

[gadav478]

Hello all. I have a calculus problem:

Find the area of the region bounded by x=y^2 and y=x-2
a)with respect to the y-axis
b)with respect to the x-axis

I understand the process but I am not sure what my professor means by with respect to x-axis.

Is it true that all we have to do is set the equations equal to y for part b and then follow the same steps for part a)?

For part a) I have:

y+2=y^2
y^2-y-2=0
Points of intersection are y=2 and y=-1
x=y^2 is the top function so we have
int from -1 to 2 (y^2 - y - 2)dy = 9/2

Do we follow the same steps in order for part b meaning:

x=y^2 will go to y=plus/minus sqrt(x)
and y=x-2 will remain the same?

Thanks in advance

 

[HallsofIvy]

Hello all. I have a calculus problem:

Find the area of the region bounded by x=y^2 and y=x-2
a)with respect to the y-axis
b)with respect to the x-axis

I understand the process but I am not sure what my professor means by with respect to x-axis.

I'm not clear on that myself! I suspect it means "integrating with respect to x" and "with respect to the y-axis" means integrating with respect to y.

Is it true that all we have to do is set the equations equal to y for part b and then follow the same steps for part a)?

For part a) I have:

y+2=y^2
y^2-y-2=0
Points of intersection are y=2 and y=-1
x=y^2 is the top function so we have
int from -1 to 2 (y^2 - y - 2)dy = 9/2

Yes, that looks good to me.

Do we follow the same steps in order for part b meaning:

x=y^2 will go to y=plus/minus sqrt(x)
and y=x-2 will remain the same?

Yes.

Thanks in advance

The two curves intersect, of course, at (4,2) and (1,-1) so integrating with respect to x you will integrate from 1 to 4. From x=0 to x= 1, The upper and lower boundaries are \(\displaystyle y= \sqrt{x}\) and \(\displaystyle y= -\sqrt{x}\). For x= 1 to 4, the lower boundary is y= x- 2 and the upper boundary is \(\displaystyle y= \sqrt{x}\)

 

[gadav478]

The two curves intersect, of course, at (4,2) and (1,-1) so integrating with respect to x you will integrate from 1 to 4. From x=0 to x= 1, The upper and lower boundaries are \(\displaystyle y= \sqrt{x}\) and \(\displaystyle y= -\sqrt{x}\). For x= 1 to 4, the lower boundary is y= x- 2 and the upper boundary is \(\displaystyle y= \sqrt{x}\)

Thanks for your help. I have another question. I don't quite see how or why we calculate boundaries here...

What does x=0 to x=1 signify and why are there different boundaries for x=1 to x=4?

Thanks again, I appreciate the help greatly.

EDIT: I think I see it. From the point of intersection, there is a change in the lower boundary because from 0 to 1 we have the (-sqrt(x)) and then it changes to the straight line (y=x-2) from 1 to 4... I had to draw it.