Can someone help me with these last three calculus problems? I'm working on finding the volume of a solid that is formed by revolving the region by the x or y axis.

Question

tkhunny · Answer

Please put enough information in the VERY FIRST post.  This includes your best efforts at a solution.  Many people see nothign in the first post and never come back.

lifeisadangerousgame · Answer

attachment

lifeisadangerousgame · Answer

@tkhunny Sorry, I'll edit it

arihangdu · Answer

Uh no questions how to solve?

arihangdu · Answer

Uh no questions how to solve?

tkhunny · Answer

Sorry, disconnected...

Why are you struggling with this?  What have you tried?

$\pi r^{2}h$ -- The area of a circle.

Differential with respect to h, $\pi r^2\;dh$

Add them all up.  $\int\limits_{0}^{4} \pi r^{2}\;dh = \int\limits_{0}^{4} \pi (\sqrt{x})^{2}\;dx$

lifeisadangerousgame · Answer

Oh I see. I went about it the wrong way, I had tried it in the same way you would for finding the area underneath a curve.

tkhunny · Answer

It's about the same.

tkhunny · Answer

Second one is nearly identical.  Set it up!

lifeisadangerousgame · Answer

After I do (sqrtx)^2 dx, and I get just x dx, do I get the anti derivative and go from there?

tkhunny · Answer

It's an integral.  How else do you plan to evaluate it?  Go!  You already know how to proceed.  No more hesitating.

tkhunny · Answer

Third one is ALMOST the same.  Just subtract the missing piece next to the y-axis.  $\pi 1^{2} 4 = 4\pi$.  You'll be done in no time.

lifeisadangerousgame · Answer

8pi for the first one?

tkhunny · Answer

Not a clue.  I didn't work the problem.  Are you confident?

lifeisadangerousgame · Answer

Not in the least

lifeisadangerousgame · Answer

But I guess it don't really mind by now, I just want to finish everything

tkhunny · Answer

Not good enough.  If you do nothing incorrectly on purpose, why would it not be correct?

lifeisadangerousgame · Answer

That's true. It should be correct then

arihangdu · Answer

👍🏻 job is done.