Find the volume of solid

Question

Find the volume of solid which is bounded above by the plane 􏰾x+y+z=5 and below by region

D={(x,y,z)􏰾􏰾z=2y, 4x²+z² ≤4}.

Hint: Try to figure out the bottom of solid. D is part of plane bounded by cylinder. 

Answer 1

This problem has been sitting there so long that I will solve.

For the bottom, 4x^2 + z^2 <= 4, z = 2y;

Thus, 4x^2 + (2y)^2 <= 4

4x^2 + 4y^2 <= 4

x^2 + y^2 <= 1

This is a circle centered at (0, 0) of radius 1. As z = 2y, the circle is actually an ellipse tilted on the infinite cylinder x^2 + y^2 <= 1

z = 2y on this, and, on the plane above x + y + z = 5, so z = 5 - x - y

Note that, on the ellipse, the maximum value of z is 2; Also, on the plane creating an ellipse above, the minimum value of z is 5 - x - y = 5 - √2. As 5 - √2 > 2, this ellipse will be above the other ellipse

Thus, z ranges from 2y to 5 - x - y, for a difference of 5 - x - 3y (we could also note that 5 - x - 3y is clearly > 1).

Thus, we integrate 5 - x - 3y over x^2 + y^2 <= 1

We could actually stop here if we wanted, as we can easily see that the average value of 5 - x - 3y on x^2 + y^2 <= 1 is 5 (for any x, y such that x^2 + y^2 <= 1, -x, -y also satisfies the equation.

Thus, as the area of the circle of radius 1 is π, the volume is 5 π

I will use I[a, b] for the integral from a to b and E[a, b] for the evaluation from a to b

Now, if you want to solve this by integration, use the polar transformation and note that r dr dθ = dx dy

As x = r cos θ and y = r sin θ

I[0, 1] I[0, 2π] r(5 - r cos θ - 3r sin θ) dθ dr =

I[0, 1] I[0, 2π] 5r - r² cos θ - 3r² sin θ dθ dr =

I[0, 1] 5r θ - r² sin θ + 3r² cos θ E[0, 2π] dr =

I[0, 1] 5r (2π) - r² sin 2π + 3r² cos 2π - (5r * 0 - r² sin 0 + 3r² cos 0) dr =

I[0, 1] 10πr - 0 + 3r² - (0 + 3r² ) dr =

I[0, 1] 10πr dr =

5πr² E[0, 1} =

5π - 0 =

5π