Finding volumes by rotating around an axis of revolution

[Sidthewall]

1. Homework Statement

k so here is the equation i need help with that will find me the volume of a sphere
2*pi*y*sqrt(25-(y-1)^2) dy - 5*pi from 0 to 6

the 5 pi is the volume of a cylinder


2. Steps
so first i subbed u=y-1
took the 2 pi out of the integral
that got me 2 integrals u*sqrt(25-(u)^2) du + sqrt(25-(u)^2) du
the first integral = (2*pi*24^(3/2))/3

the second i used a trig sub and my final answer is
(25pi^2)/2 +
-25*asin(-.2)+sqrt(24)



3. Answer

the answer is suppose to be 25pi^2 + 500pi/3

i got 25pi^2 + (-75*asin(-.2)+3*sqrt(24)+2*(24)^(3/2) -15)/3
which is approx
25pi^2 + 499.89pi/3

how do i get the exact answer

 

[HallsofIvy]

You are trying to find the volume of a sphere of radius 5 with center at (0, 1, 0)? Projected onto a x, y plane, gives the circle x^2+ (y- 1)^2= 25 and taking only the right half-plane, x= \sqrt{25- (y-1)^2}. That is the figure being rotated around the line y= 1.

And, from the way you are doing this, I take it you are using "cylinders" (I would have been inclined to use disks). Each "cylinder" of radius y has length
\sqrt{25- (y- 1)^2}- (-\sqrt{25- (y-1)^2})= 2\sqrt{25- (y- 1)^2}

and so surface area \pi r^2h= \pi (y- 1)^2(2\sqrt{25- (y-1)^2}). With "thickness" dy, each cylinder will have volume 2\pi (y- 1)^2\sqrt{25- (y-1)^2} dy so the volume of the entire sphere will be the integral, as y goes from 1 to 1+ 5= 6,
2\pi \int_1^6 (y- 1)^2\sqrt{25- (y- 1)^2}dy

A good first step, just as you say, would to be let u= y- 1 so that du= dy, when y= 1, u= 0, when y= 6, u= 5 and the integral becomes
2\pi\int_0^5 u^2\sqrt{25- u^2}du

Yes, you can let u= 5sin(\theta) so that du= 5 cos(\theta)d\theta and \sqrt{25- u^2}= \sqrt{25- 25sin^2(\theta)}= 5\sqrt{cos^2(\theta)}= 5 cos(\theta). Further, when u= 0, 5sin(\theta)= 0 so \theta= 0 and when u= 5, 5 sin(\theta)= 5 so \theta= \pi/2.

The integral becomes
2\pi\int_0^{\pi/2}(25 sin^2(\theta))(5 cos(\theta))(5 cos(\theta))d\theta[/itex]<br /> <br /> I do not know why you are using &quot;y&quot; outside the square root rather than &quot;y- 1&quot; nor do I see why it is not squared. I had to guess at what you really intended since you did not say what the radius or center of the sphere were. I also do not know why you subtracted 5\pi, the &quot;volume of a cylinder&quot; when you said you were finding the volume of a sphere.

 

[Sidthewall]

no no... sorry the circle is being revolved around the x-axis and y2= 1 which is why so the 5pi exists because i am subtracting the volume from y2

 

[Sidthewall]

oh and yes amd the question asked me to revolve the circle around the x-axis in terms of y, which is longer by the way.

 

[Sidthewall]

<br /> 2\pi \int_0^6 y\sqrt{25- (y- 1)^2}dy + 5\pi<br />

that's the integral I want solved and the answer i got is (25pi^2)/2 + (500*pi)/3 intstead of 25pi^2 + 499.89pi/3