Question

HELP I HAVE BEEN STUCK ON THIS PROBLEM FOR HOURS!!!!!!!!
Find the volume of the solid whose base is the region bounded by y=x^4, y=1 and the y-axis and whose cross-sections perpendicular to the x-axis are semicircles.
b. what would it be if the cross-sections were perpendicular to the y-axis were equilateral triangles?

Answer 1

a) Please observe symmetries. Do half the problem and multiply by 2 to finish.

For each semi-circlular cross section, we have:

Diameter = 1 - x^4

Radius = ½(Diameter) = ½(1 - x^4)

Area = \(\pi\cdot\dfrac{\left(\dfrac{1-x^{4}}{2}\right)^{2}}{2}\)

Answer 2

Really, just build it a piece at a time.

Answer 3

so then would you integrate that and evaluate it from 0 to 1?

Answer 4

Yes. And then multiply by 2.

Let's see you do the other direction.

Answer 5

Volume= integral from 0 to 1 of (1/2)(1-x^4)

Answer 6

What's that? ½(1-x^4) is the Radius of a Circle. It is not the area of a circle and it certainly is not the area of a semi-circle. Look long and hard at the expression I posted above.

Answer 7

It would be the area for an equilateral triangle that we would have to find in b right?

Answer 8

Oh, I missed that they were triangles the toehr way. I was thinking semi-circles both ways. Hold on...

Answer 9

On part a. I got 2pi(4/45) but that was the incorrect answer. I am not sure what I did wrong on it.

Answer 10

Okay, I still don't see what you have.

The BASE of the triangle is 2x. Do you see this?

So the area is (1/2)(2x)*height = x*height

Answer 11

a) 8pi/45 should be good.

Answer 12

the height of the triangle then would be sqrt3x right?

Answer 13

on a it is still saying that 8pi/45 is incorrect

Answer 14

b) Yes.
a) I see. I missed the "Bounded by the y-axis". Let's go with 4pi/45. Don't multiply it by 2.

Answer 15

now it is right in part a. thanks.

Answer 16

Same problem with b). The base is x and half the base is x/2.

Answer 17

On part b would you do the integral from 0 to 1 of (1/2)(x^1/4)(sqrt3^1/4)?

Answer 18

Missing an 'x' in there, somewhere.

Answer 19

And, no, it's just 'x', not x^(1/4).

x = y^{1/4}, though.

Answer 20

so it would bethe integral x(sqrtx^1/4) evaluated from 0 to 1?

Answer 21

No, these are perpendicular to the y-axis. You must do the integration w.r.t y.

\(\int\limits_{0}^{1}\dfrac{1}{2}x\sqrt{3}\dfrac{x}{2}\;dy = \int\limits_{0}^{1}\dfrac{1}{2}y^{1/4}\sqrt{3}\dfrac{y^{1/4}}{2}\;dy\)

Answer 22

\( = \int\limits_{0}^{1}\dfrac{\sqrt{3}}{4}\sqrt{y}\;dy\)

Answer 23

1/sqrt3?

Answer 24

I get \(\dfrac{\sqrt{3}}{6}\)

Answer 25

I see what I did wrong now.
Thank you so much for your patience and help. I really appreciate it!

Answer 26

Good work. Way to hang in there!!

Answer 27

Thanks, I couldn't have done it without you. :)