Question

Calculate the volume of S, where S is the solid which is bounded by the graph of z = x^2+1 and the planes z = 2y-1 and y = 2.

I'm having trouble calculating the bounds of x. I believe that y goes from 0 to 2 and z goes from 0 to 2y-1, but I cannot figure out x. Please help!

Answer 1

oh boy... this is hard to visualize...
i am not totally sure, but i'll have a go

Answer 2

the lower surface of the region is comprised of z=x^2 +1
and the upper by z=2y-1
so the limits for z are from 2y-1 to x^2 + 1

i feel that the projection of the required region on the x-y plane is a rectangle

so the y bounds would be:
y co-ordinate when the plane z=2y-1 hits z=x^2+1at its lowest point,
which is z=1
=>y=1

and
y=2 ( because of the vertical plane y=2)

Answer 3

i.e
y limits are from 1 to 2

Answer 4

for the x limits,
consider the parabola formed at the intersection of y=2 and z=x^2+1

the required x co-ordinates would be the furthest and nearest points along x, where the the plane z=2y-1 cuts this parabola

i.e
y=2
=> z=2(2)-1 = x^2 +1
3=x^2 + 1
\(x=\pm \sqrt{2}\)

Answer 5

edit: i have reversed the z limits

Answer 6

ok... now i think projection on xy plane might be a parabola :/
@ganeshie8 help!

Answer 7

looks it is easier to setup the projection in xz plane...

Answer 8

https://www.wolframalpha.com/input/?i= \int_1^3+\int_%28-sqrt%28z-1%29%29^%28sqrt%28z-1%29%29+\int_%28%28z%2B1%29%2F2%29^2+1+dydxdz

Answer 9

\(\href{https:///www.wolframalpha.com/input/?i=\int_1^3+\int_%28-sqrt%28z-1%29%29^%28sqrt%28z-1%29%29+\int_%28%28z%2B1%29%2F2%29^2+1+dydxdz}{here}\)

Answer 10

nice!