# Integrations Volume Question Need Help! :( (1 Viewer)

#### kevinsta

##### Member
Find the volume of the solid of
revolution formed if the area
enclosed between the curves
y=x^2 and y=(x-2)^2 is rotated

#### FrankXie

##### Active Member
there is no region bounded by these two parabolas, the correct way to refer the area is "enclosede by parabolas ... and ... and the x axis"
the solution is easy, split the region into two parts: 0 to 1, and 1 to 2. integrate respectively and add up.

#### kevinsta

##### Member
there is no region bounded by these two parabolas, the correct way to refer the area is "enclosede by parabolas ... and ... and the x axis"
the solution is easy, split the region into two parts: 0 to 1, and 1 to 2. integrate respectively and add up.
I keep getting the wrong answer tho. My algebra is clearly wrong can you perhaps show me?

#### snowflakeptical

##### ❀
^ oh yeah like this

I'll try that as well

#### snowflakeptical

##### ❀
I got 33/5 units^3
did anyone else?

#### Silly Sausage

##### Well-Known Member
Recall $\bg_white V=\pi\int^{a}_{b}{y^{2}}\, dy$
You can integrate separately for the corresponding areas by letting:
$\bg_white y^2=x^4=\pi\int^{1}_{0}{y^{2}_1}\, dy$
$\bg_white y^2=[(x-2)^2]^2=\pi\int^{2}_{1}{y^{2}_1}\, dy$
$\bg_white \pi\int^{1}_{0}{y^{2}_1}\, dy+\pi\int^{2}_{1}{y^{2}_2}\, dy$

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#### kevinsta

##### Member
I got 33/5 units^3
did anyone else?
YES I GOT THAT TOOO But the answer says 2PI/3 units^3

and I got 33Pi/5 Units^3

#### kevinsta

##### Member
Recall $\bg_white V=\pi\int^{a}_{b}{y^{2}}\, dy$
You can integrate separately for the corresponding areas by letting:
$\bg_white y^2=x^4=\pi\int^{1}_{0}{y^{2}_1}\, dy$
$\bg_white y^2=[(x-2)^2]^2=\pi\int^{2}_{1}{y^{2}_1}\, dy$
$\bg_white \pi\int^{1}_{0}{y^{2}_1}\, dy+\pi\int^{1}_{0}{y^{2}_2}\, dy$
Don't you have to combine the equations together??

#### snowflakeptical

##### ❀
YES I GOT THAT TOOO But the answer says 2PI/3 units^3

and I got 33Pi/5 Units^3
oh yeah remember how for volume it's pi outside so times the ans by pi
but why only 2 pi/3 lol?

#### kevinsta

##### Member
oh yeah remember how for volume it's pi outside so times the ans by pi
but why only 2 pi/3 lol?
My friend did it he basically just integrated y^2=x^4 and multiplied it by 2.... which makes sense when you draw the graph but like...

#### snowflakeptical

##### ❀
My friend did it he basically just integrated y^2=x^4 and multiplied it by 2.... which makes sense when you draw the graph but like...
ahhh symmetrical, which is funny because I actually posted the diagram. Dammit, but wouldn't there be more than just that method?
ok that's solved

#### kevinsta

##### Member
ahhh symmetrical, which is funny because I actually posted the diagram. Dammit, but wouldn't there be more than just that method?
ok that's solved
Yea I know I want to learn the algebraically method that always works, because usually they aren't symmetrical and I don't want to repeat the same mistake.

#### snowflakeptical

##### ❀
Yea I know I want to learn the algebraically method that always works, because usually they aren't symmetrical and I don't want to repeat the same mistake.
most of the time I sketch but omg when I do a mistake in the diagram, I'm doomed

#### kevinsta

##### Member
most of the time I sketch but omg when I do a mistake in the diagram, I'm doomed
LOOOOOOOOOOOOOOOOOL I hate drawing graphs TBH I suck at drawing Teacher tells me off when I don't draw graphs and im like IT DON'T TELL ME TO DRAW GRAPHS PLEASE... haven't met alot of people that actually use the word "doomed" like myself

#### snowflakeptical

##### ❀
LOOOOOOOOOOOOOOOOOL I hate drawing graphs TBH I suck at drawing Teacher tells me off when I don't draw graphs and im like IT DON'T TELL ME TO DRAW GRAPHS PLEASE... haven't met alot of people that actually use the word "doomed" like myself
Yes, doomed is a word used when I exaggerate a lot. My teacher likes a sketch no matter how messy because he can't draw too. Btw, you doing hsc this year?

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