Find volume of the area bounded by the two functions around the y-axis.

G

GatorChil

New member
Joined
Feb 9, 2015
Messages
4
I am having problems figuring out what the volume of this solid would be if I revolved it around the y-axis. The two functions are, y=x^2, and y= x+6. I want to find out the area enclosed by the two functions revolved around the y-axis. I found out where the two functions intersect, and I believe all I need to do now is apply the shell method, with my bounds of [0,3] and my integrand as x(x+6-x^2). I do not believe I need to account for the area bounded by the function on the left hand side, because the function's area on the right should overlap it when it is revolved around the axis, right?
 
GatorChil said:
I am having problems figuring out what the volume of this solid would be if I revolved it around the y-axis. The two functions are, y=x^2, and y= x+6. I want to find out the area enclosed by the two functions revolved around the y-axis. I found out where the two functions intersect, and I believe all I need to do now is apply the shell method, with my bounds of [0,3] and my integrand as x(x+6-x^2). I do not believe I need to account for the area bounded by the function on the left hand side, because the function's area on the right should overlap it when it is revolved around the axis, right?
Click to expand...
I want to find out the area enclosed by the two functions revolved around the y-axis. No, you want to find out the volume enclosed by the two functions revolved around the y-axis. The formula for the shell method is NOT integrate the (top function - bottom function) as you stated. That would just find the area between the two function if you have the correct limits. So what is the formula for the shell method and what do you think the integrand should be. Please respond back and you'll receive additional help.
 
GatorChil said:
I believe all I need to do now is apply the shell method, with my bounds of [0,3] and my integrand as x(x+6-x^2).
Click to expand...

Looks good; you already moved the constant 2*pi outside of the integrand, yes?


I do not believe I need to account for the area bounded by the function on the left hand side, because the function's area on the right should overlap it when it is revolved around the axis, right?
Click to expand...

You're talking about ignoring the region in Quadrant II, I think. If so, you're correct -- you may revolve the region in Quadrant I only.
 
Jomo said:
I want to find out the area enclosed by the two functions revolved around the y-axis. No, you want to find out the volume enclosed by the two functions revolved around the y-axis. The formula for the shell method is NOT integrate the (top function - bottom function) as you stated. That would just find the area between the two function if you have the correct limits. So what is the formula for the shell method and what do you think the integrand should be. Please respond back and you'll receive additional help.
Click to expand...

Oops, my mistake. Yes I want to find the volume of the area enclosed by the two functions revolved around the y-axis. I believe my bounds should be [0,3], my integrand should be 2pi(x)(x+6-x^2).
 
Quaid said:
Looks good; you already moved the constant 2*pi outside of the integrand, yes?




You're talking about ignoring the region in Quadrant II, I think. If so, you're correct -- you may revolve the region in Quadrant I only.
Click to expand...

Yes I brought out the constant. I am ignoring the area in Quadrant II, solely because the area in Quadrant I should pick up the area in Quadrant II when it revolves around the y-axis, right? Sorry if I am not explaining myself very well, but I am having trouble setting up the integral to rotate the entire area enclosed by these two functions, around the y-axis, to find this shapes volume.
 
Last edited:
GatorChil said:
Yes I brought out the constant. I am ignoring the area in Quadrant II, solely because the area in Quadrant I should pick up the area in Quadrant II when it revolves around the y-axis, right? Sorry if I am not explaining myself very well, but I am having trouble setting up the integral to rotate the entire area enclosed by these two functions, around the y-axis, to find this shapes volume.
Click to expand...
Oops I did not notice the x in front of (x+6-x^2) but I still would have complained about the missing 2pi. You have everything correct so now just evaluate the integral.
 
Jomo said:
Oops I did not notice the x in front of (x+6-x^2) but I still would have complained about the missing 2pi. You have everything correct so now just evaluate the integral.
Click to expand...

Ok great, so evaluating this integral on that interval will include the volume from the area in Quad II if I would have revolved that around the y-axis as well?
 
GatorChil said:
Ok great, so evaluating this integral on that interval will include the volume from the area in Quad II if I would have revolved that around the y-axis as well?
Click to expand...
Yes. Just look at the graph of the two functions and you'll see that this is the case.
 
You must log in or register to reply here.
Top