Suppose we have a solid S that is bounded by the following surfaces:
z = 5x^2 + 5y^2 and z = 28 - 2x^2 - 2y^2
1.) Determine the surface area of S.
2.) Determine the avg. distance from the z-axis to a point in S.
Could someone please check my work!
5x^2 + 5y^2 = 28 - 2x^2 - 2y^2
7x^2 + 7y^2 = 28
y = +/- sqrt(4-x^2) . . . z = 20
z = 5x^2 + 5y^2
f_x = 10x = 10*r*cos(theta)
f_y = 10y = 10*r*sin(theta)
z = 28 - 2x^2 - 2y^2
f_x = -4x = -4*r*cos(theta)
f_y = -4y = -4*r*sin(theta)
SA = int(int_R(sqrt(f_(x)^2 = f_(y)^2 + 1)dA
int(int(sqrt(16r^2*cos^2(theta) + 16r^2*sin^2(theta) + 1)r dr d(theta)..
Note the limits of integration are, respectively, 0 to 2Pi, 0 to 2..
I did it out and got 172.6 ...
PLEASE would someone help me determine if this is right.
For the avg. distance I'm not sure.
The second curve is parabolid which opens down. And First curve is paraboloid which open up. Their intersection forms a surface. Now the second one is one top and first one is below.
And how do they intersect? Note:
z=5x^2+5y^2
z=28-2x^2-2y^2
Thus,
5x^2+5y^2 = 28 - 2x^2 - 2y^2
Thus,
7x^2+7y^2 = 28
Thus,
x^2 +y^2 = 4
That is they intersect in a circle centered at origin or radius 2.
Now follow attachment.