A water tank is formed by revolving the curve y=3x^4 about the y-axis.?

Question

Find volume of water in tank as a function of the water depth, y.

This is a review question for a calculus exam that I've been working on for over an hour. Please help! with step by step for solving it so I can study the process before the test in 3 hours.

Answer 1

y=3x^4

so, dy=12x^3.dx

At a height (depth) y, where the radius is x, consider a ifinitesimally thin disk of thickness dy, and radius x.

It's volume, dV=(Pi)x^2.dy
=12(Pi)x^5.dx
Integrating under the limits 0 and x, we have,

V = 6(Pi).x^6
Substituting x = (y/3)^(1/4), we get

V = 2sqrt(3).(Pi)y^(3/2)

Answer 2

you is unquestionably no longer waiting to, because of employing actuality different than you already know the courting of the functionality, you have a disk of endless radius, offering you with an infinite volume. i.e. 2 distinctive factors might have thoroughly distinctive volumes. think of of two rectangles. a million) astounding y=a million and x = 10,000 with entire section xy = 10,000. This revolved around the x-axis might desire to grant a volume of 10,000*pi*(a million)^2 = 10,000 pi 2) astounding y = 10,000 and x = a million with entire section xy = 10,000. This revolved around the x axis might desire to grant an entire volume of (a million)*pi*(10,000)^2 = one hundred,000,000 * pi. different than you already know the courting of the functionality of the section under the curve, that's not possible to freshen up for the quantity. As you will see from above, the two factors are equivalent, yet their volumes are incredibly distinctive. volume isn't a functionality of the section under a curve, yet particularly a functionality of how the section lies under the curve.

Answer 3

If rotation is about y axis, the volume is given by:-
V = Int (¶x²dy) where x² = root y/root3
V = Int(¶root y/root 3) dy between limits of 0 and y
V = (¶/root 3)y^(3/2)/(3/2) = ((2¶)/(3root3))y^(3/2)
This is a nasty answer to type but I think method is OK----see what you think!

Answer 4

Normally, the problem should give you limits of integration, like the container is 6 feet tall. But the guy before me has it right. That's a good answer.