By Region

1. There is a line through the origin that divides the region bounded by the parabola. What is the slope?

Find the zeros of the function to find the interval of integration. y = 2x - 6x² = 2x(1 - 3x) = 0 x = 0, 1/3 The area between the curve and the x-axis is: ∫(2x - 6x²)dx = x² - 2x³ | [Evaluated from 0 to 1/3] = (1/9 - 2/27) - 0 = 1/27 __________ Now find where the parabola intersects with the line y = mx. y = mx = 2x - 6x² 6x² + (m - 2)x = 0 x[6x - (2 - m)] = 0 x = 0, (2 - m)/6 Integrate the area enclosed by the parabola and the line. ∫[(2x - 6x²) - mx] dx = ∫[(2 - m)x - 6x²] dx = (2 - m)x²/2 - 2x³ | [Evaluated from 0 to (2 - m)/6] = (2 - m)³/(2*6²) - 2[(2 - m)/6]³ - 0 = (1/72 - 2/216)(2 - m)³ = (1/216)(2 - m)³ = (1/2)(1/27) = 1/54 Now we have: (1/216)(2 - m)³ = 1/54 (2 - m)³ = 4 2 - m = 4^(1/3) m = 2 - 4^(1/3) ≈ 0.4125989

2. How do I shade a region bounded by two functions on a TI-83 calculator from a program?

u have to have another y = o

3. Find the volume of the solid formed by rotating the region enclosed by?

dV = [π(3 + x^7)^2]dx dV = π(9 + 6x^7 + x^14)dx V = π[9x + (3/4)x^8 + (1/15)x^15] | from x=0 to x=1 V = π[9(1 - 0) + (3/4)(1 - 0) + (1/15)(1 - 0)] V = π(540 + 45 + 4)/60 V = π(589/60) ≈ 30.83997 cubic units

4. Find the volume of the solid generated by revolving the region in the first quadrant?

Problems like this are usually solved using what is sometimes called 'the method of shells'. The formula for this volume is: V = 2 pi int x f(x) dx In this case, f(x) = sin 2x and the integral goes from 0 to pi/2. Therefore: V = 2 pi int x sin 2x dx We can integrate this by parts. The indefinite integral is: int x sin 2x dx = x (-1/2 cos 2x ) - int (-1/2 cos 2x ) dx = -( x cos 2x ) / 2 + 1/2 ( 1/2 sin 2x ) = ( sin 2x - 2x cos 2x ) / 4 V = 2 pi [ ( sin 2x - 2x cos 2x ) / 4 ] (0, pi/2) = 2 pi ( (0 - pi*(-1))/4 - (0 - 0)/4 ) = pi^2/2 Therefore, the volume is pi^2 / 2 = 4.93.

5. What is the volume of the solid of revolution formed by rotating about the x-axis the region between the graph?

6. Does anywhere know where to find Samuel Adams sales by region?

Did you try contacting the company for that? Also, try their investor's site, for the annual report. They're public so they have to disclose some sort of information, whether they break it down by region or not is up to them.

7. How is the design influenced by the region and period of time?

definitely, designs are influenced by the region.. in relation to their culture.. resources available in the area.. the climate.. domestic preference.. and so on.. and yes, equally dictated by time.. history of the region.. restoration.. preservation.. present times.. and future designs are based on all this regional factors

8. Find the volume of the solid of revolution obtained by rotating the region bounded by the curves y=x^2?

First you find intersection is at x=1 and x=-1, but since you find volume of rotation about y-axis, then you integrate from 1 to 0. So, for y=x^2, V=2pi INT x*y=2pi INT x^3 from x=1 to 0= 2pi x^4/4 from x=1 to 0 = 2pi. And for y=2-x^2, V=2pi INT x*y = 2pi INT 2x-x^3 = 2pi INT x^2-x^4/4 from x=1 to 0 = 2pi (1-(1/4)) = 2pi(3/4). So, total V= 2pi (1-(3/4)) = 2pi(1/4) = pi/2, I think.

9. Find the area of the region bounded by the parabola?

Edit: I messed up the limits of integration on the tangent line. It should be (1,2) not (0,2) First you need to find the tangent line. remember that any line is of the form y=mx+b The slope of the tangent line to a graph at a given x is defined as the derivative of the graph evaluated at that x. So to find the slope of the tangent line of y=3x^2, take the derivative, to get y'=6x, then evaluate at 2 y'(2)=6(2) y'(2)=12 so the line is y=12x+b to find b plug in the known point (2,12) and solve. 12=12(2)+b 12=24+b -12=b so the tangent line is y=12x-12 when you plot the graph and its tangent line, you notice that the area your looking for is the area between the parabola and the x-axis minus the area between the tangent line and the x-axis. so take the integral of the parabola (from 0 to 2) minus the integral of the tangent line (from 1 to 2). (The boundries should be obvious once you graph it as well.) so Int(3x^2) eval 0,2 minus Int(12x-12) eval1,2 x^3 (0,2) minus 6x^2-12x (1,2) (8-0) minus (0--6) 8-6 2.

10. What is the volume of this solid generated by rotating the region about the x-axis?

The figure you get is a cylinder with a cone shaped hole in at at one end. The base of the cone is the same as the cylinder's base. Dimensions of the cylinder: radius = 1 and height = 1 Dimensions of the cone: radius = 1 and height = 1 Required volume = Volume of cylinder - Volume of cone = pi * (1) ^ 2 * (1) - 1/3 * pi * (1) ^ 2 * (1) = ( 2/3 ) * pi Hope this helps...

11. The region enclosed by the line x+y=1 and the coordinate axes is rotated about the line y=-1. What is volume..

E. 4π/3 Volume = ∫(x=0 to x=1) π[(y+1)² - 1] dx = π ∫(x=0 to x=1) (2-x)² -1] dx = π ∫(x=0 to x=1) (3 - 4x + x²) dx = π (3x - 2x² + x³/3) (x=0 to x=1) = π (3 - 2 + 1/3) = 4π/3.

12. Is the route 129a going to be replaced by york region transit if ttc go on strike?

Nope ... you'd be SOL ...

13. Find the volume of the region in the first octant that is bounded by the hyperbolic cylinders?

The Jacobian determinant for the suggested transformation is: |d(u,v,w)/d(x,y,z)| = |y x 0 , z 0 x , 0 z y| = -2xyz == dudvdw = dxdydz(-2xyz) == dxdydz = -dudvdw/(2xyz) = -dudvdw/(2*sqrt(uvw)) (The sign just depends on the order of variables and is not significant.) All variables are positive. The volume is the integral: \int_1^9 du \int_1^4 dv \int_1^16 dw 1/(2*sqrt(uvw) = (1/2) * \int_1^9 du/sqrt(u) * \int_1^4 dv/sqrt(v) * \int_1^16 dw/sqrt(w) =(1/2) * 2^3 * (3-1)*(2-1)*(4-1) = 24

14. Can you record WinUAE in windowed mode by selecting region in CamStudio?

Use Bulent's Screen Recorder : http://www.thesilver.net/downloadbsr.asp Check 'Disable Video Acceleration When Recording' option in Configurations-Page I-Options. This should prevent blank recording.

15. how do i find volume of solid generated by revolving the region?

Volume=Integral from A to B of A(x)dx or A(y) dy where: A=pi(radius)^2 if cross section is a disk A=pi(outer radius)^2-pi(inner radius)^2 if cross section is a washer Just integrate that using the bounds given in the problem (a=x=o and b=x=sq. root of pi) I dont envy you, good luck :)

16. What evidence shows that the Egypt was first unified by rulers from the south around the region of Nekhen?

Everything is proven in the Bible - you just have to know how to interpret it.

17. Can anyone tell me where to get information on bottled water consumption, by brand, by region?

well almost everything u buy has a website for their company and a hot line number

18. What is the area of the region enclosed by the parametric equation: x = t^3 - 7 t, y = 4 t^2?

It does form an enclosed region, try graphing with the following limits for t: -√7 ≤ t ≤ √7, this will just show the enclosed region, or try between -5 and 5, to get a rough idea of what happens afterwards. Then just find the enclosed area, which I figure you know how to do.

19. Find the volume generated by rotating the region bounded by?

y1=e^x and y2=π intersect at y1=y2 == e^x=π == x=lnπ V=π∫[0, lnπ] (y2^2-y1^2) dx = π∫[0, lnπ] [π^2-(e^x)^2)] dx = π∫[0, lnπ] [π^2-e^(2x)] dx = π[xπ^2-0.5e^(2x)] [x=0 to lnπ] = π[(π^2)lnπ-0.5e^(2lnπ)+1/2]= π[(π^2)lnπ-0.5π^2+1/2] ~ 21.56

20. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified a

/ v=pi int from y=0 to 1 of 25y^4dy v=pi(5y^5) from 0 to 1 =5pi