(a) Consider the region in the xy-plane consisting of the points (x; y) satisfying x > 0, y > 0, and lying between the curves y = x^2 + 1, y = 2x^2 2 and the two axes. Draw
a diagram of this region.
(b) This region is rotated about the y-axis to form a solid glass vase. Determine the
volume of glass in the vase.
3 answers
y=x^2+1, y=2x^2-2 ***
Here is the sketch, the bottom diagram shows it best
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+%2B+1%2C+y+%3D+2x%5E2+-+2
We need their intersection:
2x^2 - 2 = x^2 + 1
x^2 = 3
x = ± √3 , y = 4
I will do it in two parts,
1. horizontal discs from y = -2 to y = 1, where I will use y = 2x^2 - 2
2x^2 = y+2
x^2 = y/2 + 1
Volume = π∫(y/2 + 1)dy/ from y = -2 to 1
= π[(1/4)y^2 + y] from -2 to 1
= π(1/4 + 1 - (4 - 4)
= 5π/4
2. now for the top part:
V = π∫(y/2 + 1 - (y - 1) dy/ from 1 to 4
= π∫( y/2 - y + 2) dy from 1 to 4
= π[y^2 /4 - y^2 /2 + 2y] from 1 to 4
= π( 4 - 8 + 8 - ( 1/4 - 1 + 2))
= π(4 - 5/4)
= 21π/4
total volume = 13π/2
check my arithmetic, should have written it out on paper first.
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2+%2B+1%2C+y+%3D+2x%5E2+-+2
We need their intersection:
2x^2 - 2 = x^2 + 1
x^2 = 3
x = ± √3 , y = 4
I will do it in two parts,
1. horizontal discs from y = -2 to y = 1, where I will use y = 2x^2 - 2
2x^2 = y+2
x^2 = y/2 + 1
Volume = π∫(y/2 + 1)dy/ from y = -2 to 1
= π[(1/4)y^2 + y] from -2 to 1
= π(1/4 + 1 - (4 - 4)
= 5π/4
2. now for the top part:
V = π∫(y/2 + 1 - (y - 1) dy/ from 1 to 4
= π∫( y/2 - y + 2) dy from 1 to 4
= π[y^2 /4 - y^2 /2 + 2y] from 1 to 4
= π( 4 - 8 + 8 - ( 1/4 - 1 + 2))
= π(4 - 5/4)
= 21π/4
total volume = 13π/2
check my arithmetic, should have written it out on paper first.
1. I get
= π[(1/4)y^2 + y] from -2 to 1
= π((1/4+1)-(1-2))
= 9π/4
2. I get 9π/4 also
making the total volume 9π/2
= π[(1/4)y^2 + y] from -2 to 1
= π((1/4+1)-(1-2))
= 9π/4
2. I get 9π/4 also
making the total volume 9π/2