Volume of Solids of Revolution

[misty pollen]

In a recent quiz about including getting the volume of solids of revolution, this problem came up:

The plane region bounded by y=sqrt(x) and 2=x+y is revolved about y=1. Determine the volume of the solid generated. Use the circular Shell Method.

Isn't the problem incorrectly asked? I get what my professor is trying to make us do with the original curves they gave, which is to transform y=sqrt(x) to x=y^2; however, doesn't that make the math "wrong?"

Shouldn't my professor have explicitly stated that the volume/solid generated was bounded by the curve x=y^2 and not what they gave (y=sqrt(x))?

[visual summit]

👋 Welcome to your Help Thread!

Hey @misty pollen, Thanks for sharing your math problem with us. While you wait for a Helper to help you, we want to share some vital information with you.

●

Please take a moment to read the helpee guidelines. This will make sure that your post follows the helpee rules of our community.

●

Please don't ping <@&775784618955505685>, <@&1283689826742440016>, <@&819616364188139550>, or <@&624327278137966593> for help because their job is to take care of the server's administrative tasks, not to answer queries directly. However, if you have a problem with how a Helper is acting, you can ping a Helper Moderator.

●

It's always very useful if you can show us the work you've done so far. This makes it easier for our Helpers to find mistakes and help you get to the right answer.

Once you've got your answer and you're all set, you can close this thread by typing +close, and then you can click the Thank buttons to show your Helper some appreciation. · May 24, 2026, 10:23 AM

[misty pollen]

What I meant by "wrong" is that x=y^2 and y=sqrt(x) produce two different graphs and transforming either of the two introduces fake solutions. Is my understanding correct?

[mellow comet]

yes, it changes the graph globally
but no, your previous range was [0, 1]. That is unaffected by squaring sides. Only the negative range is created
Do you think that makes previous process "wrong"?

[misty pollen]

that's where my confusion is... squaring both sides introduces the negative range of y=sqrt(x), but those added points were never part of the original function

the original range [1,0] is still valid, but the transformed equation now introduces extra intersections that are impossible under the original constraints y>=0

for this problem to be possible to solve, shouldn't the problem have stated that only the range of y=0 to y=1 was to be considered since, again, the negative range was only introduced as a result of manipulating the original function. not an actual solution.

this is the solution my professor wants btw

[mellow comet]

Hmm, so he meant to include the fake portion?

Then I will certainly agree with you
The problem should have been stated more clearly

[misty pollen]

thanks a lot!

[visual summit]

@misty pollen

<:HelpIcon:1304095958283321385>| Help Reminder

Hello .gab8266, this is a friendly reminder that your help request has been inactive for more than 24 hours. If you no longer need assistance, please consider closing the thread using the +close command. This thread will be automatically closed in 3 days if it remains inactive.

[misty pollen]

+close

[visual summit]

Please thank your Helpers before closing!

Please thank the helpers who assisted you by clicking the buttons below. You can thank each helper only once. Once you're done, click "Close Post" to close this thread.

[visual summit]

Thank you for your feedback! Xer0 has been awarded 1 . They now have 4 . They have 3 daily left for today.