a) Find the volume formed by rotating the region enclosed by x = 6y and y^3 = x with y greater than, equal to 0 about the y-axis.
b) Find the volume of the solid obtained by rotating the region bounded by y = 4x^2, x = 1, and y = 0 about the x-axis.
c) Find the volume of the solid obtained by rotating the region bounded by y = 1/x^6, y = 0, x = 3, and x = 9, about the x-axis.
Thanks for your trouble.
I'll do (a)
First, determine where the curves intersect:
6y = y^3 at the point (6√6,√6)
You can do this using discs or shells. Using discs, or washers, you have a stack of washers. The area of a disc of outer radius R and inner hole of radis r is
m(R^2 - r^2)
Since 6y > y^3 on the interval desired
The thickness of each disc is dy. So, the volume of the stack of discs is the integral
π Int(R^2 - r^2 dy)[0,√6]
= π Int(36y^2 - (y^3)^2) dy)[0,√6]
= π Int(36y^2 - y^6 dy)[0,√6]
= π (12y^3 - 1/7 y^7)[0,√6]
= π (12*6√6 - 1/7 * 216√6)
= 36π√6(2 - 6/7)
= 36*8π√6/7
Or, if you want to calculate using shells, the volume of a shell is 2πrh
r = x, h = x^(1/3) - x/6
So, the volume of a ring of shells is
2π Int(x(x^1/3 - x/6) dx )[0,63/2]
= 2π Int(x(x^1/3 - x/6) dx )[0,63/2]
= 2π Int(x4/3 - x2/6) dx )[0,63/2]
= 2π(3/7 x7/3 - x3/18)[0,63/2]
= 2π(67/2 - 69/2/18)
= 2π*216√6(3/7 - 1/3)
= π*216√6(4/21)
= 36*8π√6/7
Thank you!
a) Ah, the good old rotating regions! Let's kick this off. To find the volume, we'll use the method of cylindrical shells. But before we proceed, let me tell you a joke to lighten the mathematical mood:
Why did the scarecrow win an award?
Because he was outstanding in his field!
Okay, back to business. The region enclosed by x = 6y and y^3 = x looks like a fancy curve, doesn't it? We need to figure out the limits of integration, so let's solve these equations first.
From x = 6y, we find y = x/6.
Now, y^3 = x becomes (x/6)^3 = x, which simplifies to x^3/216 - x = 0. Solving this equation gives us x = 0 and x = 216.
Now, let's set up the integral for the volume:
V = ∫(2πy)(x)dy, as y ranges from 0 to 216/6 = 36.
Simplify the equation, substitute x = 6y:
V = ∫(2πy)(6y)dy, from 0 to 36.
V = 12π∫y^2dy, from 0 to 36.
V = 12π * (1/3)y^3 | from 0 to 36.
V = 12π * (1/3) * (36^3) - (1/3)(0^3).
V = 12π * (1/3) * (46656 - 0).
V = 12π * 15552.
V = 186624π cubic units.
Voila! The volume formed by rotating the region about the y-axis is 186624π cubic units.
b) Now, let's tackle the next problem. We have the region bounded by y = 4x^2, x = 1, and y = 0. Time for another joke while we're at it:
Why don't scientists trust atoms?
Because they make up everything!
Alright, let's find the limits of integration here. We can see that y ranges from 0 to 4x^2, and x ranges from 0 to 1.
To find the volume using cylindrical shells, we have:
V = ∫(2πx)(y)dx, with y = 4x^2, and x ranges from 0 to 1.
Substitute y = 4x^2:
V = ∫(2πx)(4x^2)dx, from 0 to 1.
V = 8π∫x^3dx, from 0 to 1.
V = 8π * (1/4)x^4 | from 0 to 1.
V = 8π * (1/4)(1^4 - 0^4).
V = 8π * (1/4) * 1.
V = 2π cubic units.
Hooray! The volume of the solid obtained by rotating the region about the x-axis is 2π cubic units.
c) Now, onto the last one! We have the region bounded by y = 1/x^6, y = 0, x = 3, and x = 9. Here's your final joke:
Why don't skeletons fight each other?
They don't have the guts!
Now, let's find the limits of integration. We have y ranging from 0 to 1/x^6, and x ranging from 3 to 9.
Using cylindrical shells, the volume is given by:
V = ∫(2πx)(y)dx, with y = 1/x^6, and x ranging from 3 to 9.
Substitute y = 1/x^6:
V = ∫(2πx)(1/x^6)dx, from 3 to 9.
V = 2π∫(1/x^5)dx, from 3 to 9.
V = 2π * (-1/4)x^-4 | from 3 to 9.
V = 2π * [-(1/4)(1/9^4) + (1/4)(1/3^4)].
V = 2π * [-(1/4)(1/6561) + (1/4)(1/81)].
V = 2π * [-(1/4)(1/6561) + (1/4)(1/81)].
And that's where I'll stop and let you finish the calculation. Trust me, it may look complicated, but you'll get there. Good luck and keep smiling!
a) To find the volume formed by rotating the region enclosed by x = 6y and y^3 = x about the y-axis, you can use the method of cylindrical shells.
First, let's find the intersection points between the two curves:
x = 6y
y^3 = x
Substituting x = 6y into the second equation:
y^3 = 6y
y(y^2 - 6) = 0
Therefore, y = 0 and y = ±√6 are the three intersection points.
Now, let's define the cylindrical shells that will be used to calculate the volume. Each shell will have a height equal to the difference between the two curves at a given y-value, and a radius equal to the y-value.
The radius of the shell is y, and the height is given by:
h = x_upper - x_lower
h = 6y - y^3
The volume of each shell can be calculated using the formula:
V_shell = 2πrh
Integrating the volume over the range of y values where the curves intersect, we get the total volume:
V_total = ∫(2πy)(6y - y^3) dy
V_total = 2π ∫(6y^2 - y^4) dy
Evaluating the integral:
V_total = 2π [(2y^3)/3 - (y^5)/5] evaluated from y = 0 to y = √6
b) To find the volume of the solid obtained by rotating the region bounded by y = 4x^2, x = 1, and y = 0 about the x-axis, we'll again use the method of cylindrical shells.
First, let's find the intersection points between the curves:
y = 4x^2
x = 1
y = 0
The intersection points are x = 1 and y = 0.
Now, define the cylindrical shells with radius x and height given by y:
h = y
r = x
The volume of each shell can be calculated using the formula:
V_shell = 2πrh
Integrating the volume over the range of x values where the curves intersect, we get the total volume:
V_total = ∫(2πx)(y) dx
V_total = 2π ∫(4x^2)(x) dx
Evaluating the integral:
V_total = 2π [(4/3)x^4] evaluated from x = 0 to x = 1
c) To find the volume of the solid obtained by rotating the region bounded by y = 1/x^6, y = 0, x = 3, and x = 9 about the x-axis, once again we'll use the method of cylindrical shells.
First, let's find the intersection points between the curves:
y = 1/x^6
y = 0
x = 3
x = 9
The intersection points are x = 3, x = 9, and y = 0.
Now, define the cylindrical shells with radius x and height given by y:
h = y
r = x
The volume of each shell can be calculated using the formula:
V_shell = 2πrh
Integrating the volume over the range of x values where the curves intersect, we get the total volume:
V_total = ∫(2πx)(y) dx
V_total = 2π ∫(1/x^6)(x) dx
Evaluating the integral:
V_total = 2π [(-1/5x^5)] evaluated from x = 3 to x = 9
By following these steps, you can calculate the volumes of the solids formed by rotating the given regions about the specified axes.