Find the volume of the solid that results when the region enclosed by the given curves is revolved about the..

Question

Find the volume of the solid that results when the region enclosed by the given curves is revolved about the y-axis (use "pi")

x=sqrt(1+y), x=0, y=3

volume=? Thanks a lot <|=D

Answer 1

when you sketch that you get a paraboloid
so what we need here is a tripple integral, on which we have to set the bounds.
If we denote phi as the angle, r as the radius and z as the height we can conclude the following:
Since the solid is a rotational solid that means that the integration by phi goes from 0 to 2Pi and is independent of integrations by r and z.

(1) phi goes from 0 to 2Pi

if we look at the formula x = sqrt(1+y) we can easily transform it to

(2) y = x*x - 1

for the part of the curve that participates in the region
given that we get that

(3) x goes from 0 to 2

if we slice the solid with any plane containing the z axes we get the same region as the one we started form. Therefore we can easily conclude that r changes in the same manner as x does

(4) r goes from 0 to 2

from the area enclosed by the given curves we see that the y parameter changes with x parameter using the formula y = x*x - 1
and if we take into consideration that the plane containing z axes slices the solid always in the same region we can easily conclude that z changes with r using the same formula z = r*r - 1.
Hence that yields

(5) z goes from -1 to r*r - 1

With that determined all we need is to solve the integral defined as follows (I'll use the §{low, up} for the integral in lack of better sign):

(6) V = §{0, 2Pi} §{0,2} §{-1, r*r-1} r dz dr dphi

Since phi is independent of r and z and we put r out of the integration by z we get

(7) V = 2Pi §{0, 2} r §{-1, r*r - 1} dz dr

if we solve the integral by z (which is pretty simple) (7) yields

(8) V = 2Pi §{0, 2} r(r*r - 1 + 1) = 2Pi §{0, 2} r*r*r

Solving this integral (which is pretty simple as well) we get that the volume is expressed in the function

(9) Pi * r*r*r*r/2

where r goes from 0 to 2 yielding the final result is

(10) V = 8Pi

Answer 2

that's no longer too annoying. in case you draw the diagram you will see that it incredibly is a beneficial common one, y starts off from 7 at x = 0 to 7.2^(-a million/4) at x = pi/4 and is going right down to 0 at x = pi/2. So we in simple terms opt for V = pi. int(pi/4 to pi/2) y^2 dx = pi. int(pi/4 to pi/2) 40 9 cos x dx = pi. 40 9 sin x [pi/4 to pi/2] = 40 9 pi [sin pi/2 - sin pi/4] = 40 9 pi (a million - a million/sqrt(2)) in case you opt for to rationalise the denominator you may rewrite this as = 40 9 pi (a million - sqrt(2) / 2).