Sandwiched Paraboloids

Calculus Level 2

Consider these two paraboloids:

z = x 2 + y 2 z = 2 ( x 2 + y 2 ) z = x^2 + y^2 \\ z = 2 \, (x^2 + y^2)

What is the volume enclosed between the two paraboloids, in the range 0 ≤ z ≤ 1 0 \leq z \leq 1 ?


The answer is 0.7854.

Steven Chase by Steven Chase , 37, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Tom Engelsman
May 8, 2018

Using cylindrical coordinates, the required volume computes to:

V = V p a r a b o l o i d 1 − V p a r a b o l o i d 2 = ∫ 0 2 π ∫ 0 1 ∫ r 2 1 r d z d r d θ − ∫ 0 2 π ∫ 0 1 2 ∫ 2 r 2 1 r d z d r d θ = π 4 . V = V_{paraboloid 1} - V_{paraboloid 2} = \int_{0}^{2\pi} \int_{0}^{1} \int_{r^2}^{1} r dz dr d\theta - \int_{0}^{2\pi} \int_{0}^{\frac{1}{\sqrt{2}}} \int_{2r^2}^{1} r dz dr d\theta = \boxed{\frac{\pi}{4}}.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...