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# P3 Integration watch

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1. 1) The finite region bounded by the curve with equation y=tan0.5x, the line x=pi/2 and the coordinate axes is rotated through 2pi radians about the x-axis. Show that the volume generated is [pi(4-pi)]/2.

2) By rotating the semi-circle for which y> or equal to 0 from the circle with equation x^2 + y^2 = a^2 completely about the x-axis, show that the volume of a sphere, of radius a, is [4pi(a^3)]/3.

3) A triangular region is bounded by the line with equation y = rx/h, where r and h are positive constants, the line x=r and the x-axis. By considering the complete rotation of this triangle about the x-axis, show that the volume of a cone of height h and base radius r is [pi(r^2)h]/3.

4) An ellipse is given by x=3cost, y=2sint, 0<or= t <or= 2pi.

a) Find the area of the finite region bounded by the ellipse and the positive x-and y-axes.

b) This region is rotated completely about the y-axis to form a solid of revolution. Find the volume of this solid.

4a) 3pi/2.
4b) 12pi.

Cheers
2. 1.
V = pi ∫ tan²(x/2) dx
= pi ∫ sec²(x/2) - 1 dx
= pi [2tan(x/2) - x]
= pi [(2 - pi/2) - (0 - 0)]
= pi (4 - pi)/2

2.
Find x-incercepts by setting y=0:
x = ±a
So:
V = pi ∫ a² - x² dx
= pi [a²x - (x³)/3]
= pi [(a³ - (a³)/3) - (-a³ + (a³)/3)]
= pi (4/3) a³

3.
V = pi (r/h)² ∫ x² dx
= pi (r²/3h²) [ x³ ], since the height of the cone we want is h.
= pi (r²/3h²) . h³
= (pi r² h)/3

4. Are you sure you typed it out correctly? I can't seem to get those answers.
3. (Original post by dvs)
1.
V = pi ∫ tan²(x/2) dx
= pi ∫ sec²(x/2) - 1 dx
= pi [2tan(x/2) - x]
= pi [(2 - pi/2) - (0 - 0)]
= pi (4 - pi)/2

2.
Find x-incercepts by setting y=0:
x = ±a
So:
V = pi ∫ a² - x² dx
= pi [a²x - (x³)/3]
= pi [(a³ - (a³)/3) - (-a³ + (a³)/3)]
= pi (4/3) a³

3.
V = pi (r/h)² ∫ x² dx
= pi (r²/3h²) [ x³ ], since the height of the cone we want is h.
= pi (r²/3h²) . h³
= (pi r² h)/3

4. Are you sure you typed it out correctly? I can't seem to get those answers.

I've checked it again, and I did type it out correctly
4. One question - why didn't you square the whole thing at the start of each of the 3 questions? If you want to find the volume, shouldn't you square all the contents before you integrate it?
5. (Original post by Aristotle)
...

4) An ellipse is given by x=3cost, y=2sint, 0<or= t <or= 2pi.

a) Find the area of the finite region bounded by the ellipse and the positive x-and y-axes.

b) This region is rotated completely about the y-axis to form a solid of revolution. Find the volume of this solid.

4a) 3pi/2.
4b) 12pi.

Cheers
4/
a)
A = int y dx
= int 2sint.(-3sint) dt {pi/2, 0}
= -6 int sin²t dt {pi/2, 0}
= -3 int (1 - cos2t) dt {pi/2, 0}
= -3[t - (1/2)sin2t]{pi/2, 0}
= -3{(0 - 0) - (pi/2 - 0)}
= 3pi/2
=====

b)
for this part I get
V = 8pi
=====

Anyone else ?
6. (Original post by Aristotle)
One question - why didn't you square the whole thing at the start of each of the 3 questions? If you want to find the volume, shouldn't you square all the contents before you integrate it?
dvs did "square" everything.
He used the formula

V = pi int y² dx

each time.
7. (Original post by Fermat)
b)
for this part I get
V = 8pi
=====

Anyone else ?
I get that too! A mistake in the book perhaps?
8. (Original post by dvs)
I get that too! A mistake in the book perhaps?
Obviously

4/
b)

eqn of ellipse is

(x/3)² + (y/2)² = 1
y²/4 = 1 - x²/9
y² = 4(1 - x²/9)

V = pi int y² dx
V = pi int 4(1 - x²/9) dx
V = 4pi int (1 - x²/9) dx
V = 4pi[x - x^3/27] { 0,3}
V = 4pi{(3 - 3^3/27) - (0 0 0)}
V = 4pi{3 - 1}
V = 8pi
=====

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