Finding volume between two surfaces.

[bigbk92]

I am having trouble finding the volume between the two surfaces. Well I did the problem and got an answer but I am not sure if I am approaching this type of problem correctly. So can you take a look at how i approached the problem and tell me if i am on the right path or if it is incorrect. Thank you in advance.

Find the volume between the two surfaces
z = x^2 + 3y^2
z = 8 - x^2 - y^2

First I set both equations equal to each other to get y =±sqrt(2 - x^2).
Finding the volume will require a double integral so this can be my y limits.
X cannot exceed sqrt(2) so i assume my x limits are ±sqrt(2)

The sum of the two functions is 2y^2 + 8. I believe this is what i should integrate. If not please tell me what i should integrate and why.

So i integrate 2y^2 + 8 with respect to y and get (2/3)y^3 + 8y. And inputting the y limits yields (4/3)*(2 - x^2)^(3/2) + 16*(2 - x^2)^(1/2).

Now i integrate that with respect to x using the computer and get
18*arcsin(sqrt(2)x/2) - (1/3)x*(x^2 - 5)*sqrt(2 - x^2) + 8x*sqrt(2 - x^2)

Putting in limits of ±sqrt(2) gets a final answer of 18*pi.

 

[HallsofIvy]

I am having trouble finding the volume between the two surfaces. Well I did the problem and got an answer but I am not sure if I am approaching this type of problem correctly. So can you take a look at how i approached the problem and tell me if i am on the right path or if it is incorrect. Thank you in advance.

Find the volume between the two surfaces
z = x^2 + 3y^2
z = 8 - x^2 - y^2

First I set both equations equal to each other to get y =±sqrt(2 - x^2).

Well, you shouldn't get that! Did you forget the "3" multiplying y^2 in the first equation? z= x^2+ 3y^2= 8- x^2- y^2 gives 2x^2+ 4y^2= 8
or x^2+ 2y^2= 4. That's an ellipse. Solving for y, y= \pm(1/2)\sqrt{4- x^2}.

Finding the volume will require a double integral so this can be my y limits.
X cannot exceed sqrt(2) so i assume my x limits are ±sqrt(2)

The sum of the two functions is 2y^2 + 8. I believe this is what i should integrate. If not please tell me what i should integrate and why.

If the "ceiling" is given by f(x,y) and the "floor" by g(x,y) the "height" is f- g, not f+ g. Subtract, don't add! Here, the "ceiling" is f(x,y)= 8- x^2- y^2 and the "floor" is g(x,y)= x^2+ 3y^2. The distance between them, the height, is (8- x^2- y^2)- (x^2+ 3y^2)= 8- 2x^2- 4y^2.

So I integrate 2y^2 + 8 with respect to y and get (2/3)y^3 + 8y. And inputting the y limits yields (4/3)*(2 - x^2)^(3/2) + 16*(2 - x^2)^(1/2).

Now i integrate that with respect to x using the computer and get
18*arcsin(sqrt(2)x/2) - (1/3)x*(x^2 - 5)*sqrt(2 - x^2) + 8x*sqrt(2 - x^2)

Putting in limits of ±sqrt(2) gets a final answer of 18*pi.

As above, x goes from -2 to 2 and, for each x, y goes from -(1/2)\sqrt{4- x^2} to (1/2)\sqrt{4- x^2}.
\int_{x=-2}^2\int_{-(1/2)\sqrt{4- x^2}}^{(1/2)\sqrt{4- x^2}}8- 2x^2- 4y^2 dydx

Another way to do this is to write the ellipse, at which the two paraboloids intersect, as
\frac{x^2}{4}+ \frac{y^2}{2}= 1

Let x= 2rcos(\theta), y= \sqrt{2}rsin(\theta) so that the Jacobian is
\left|\begin{array}{cc}\frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial\theta}\end{array}\right|= \left|\begin{array}{cc}2cos(t) & -2rsin(\theta) \\ \sqrt{2}sin(\theta) & \sqrt{2}rcos(\theta) \end{array}\right|= 2\sqrt{2}r
And the integrand is 8- 2x^2- 4y^2= 8- 4r^2cos^2(\theta)- 8r^2sin^2(\theta). You can use the fact that sin^2(\theta)+ cos^2(\theta) to write this as 8- 4r^2(cos^2(\theta)+ sin^2(\theta))- 4r^2sin^2(\theta)= 4- 4r^2sin^2(\theta)
so the integral would be
\int_{r=0}^1\int_{\theta= 0}^{2\pi} (4- 4r^2sin^2(\theta))(2\sqrt{2} r dr d\theta)