Question: Compute the volume of the solid bounded above and below by the surfaces z=4-x and z= (2*(x^2)) + (y^2) I found the intersection of above 2 surfaces to be the below; (x+ (1/4))^2 + (y/(sqrt(2))^2 = 33/16 How do we proceed from here?

Question

Question: Compute the volume of the solid bounded above and below by the surfaces z=4-x and z= (2*(x^2)) + (y^2) I found the intersection of above 2 surfaces to be the below; (x+ (1/4))^2 + (y/(sqrt(2))^2 = 33/16 How do we proceed from here?

Answer 1

To compute the volume of the solid bounded above and below by the surfaces z=4-x and z=(2(x^2))+(y^2), we need to integrate the function f(x,y) = 4-x - ((2x^2) + y^2) over the region R bounded by the intersection curve of the two surfaces.

As you have already found, the intersection curve is given by (x+1/4)^2 + (y/sqrt(2))^2 = 33/16. Note that this is the equation of an ellipse centered at (-1/4,0) with semi-major axis length sqrt(33)/4 and semi-minor axis length sqrt(33/2)/4.

We can use this information to set up the limits of integration. Since the ellipse is centered at (-1/4,0), we can shift the coordinates by substituting u = x+1/4 and v = y/sqrt(2). Then, the equation of the ellipse becomes u^2 + v^2 = 33/16.

To integrate over this region, we can use polar coordinates. We have u = rcos(theta) and v = rsin(theta), and the limits of integration become:

0 ≤ r ≤ √(33)/4
0 ≤ theta ≤ 2π

Thus, the volume of the solid is given by:

V = ∫∫R (4-x - ((2*x^2) + y^2)) dA
= ∫∫R (4 - (2*x^2) - y^2 - x) dA
= ∫0^2π ∫0^(√(33)/4) (4 - 2(r^2cos^2(theta)) - r^2sin^2(theta) - rcos(theta)) r dr dtheta

Simplifying this expression, we get:

V = ∫0^2π ∫0^(√(33)/4) (4r - 2r^3cos^2(theta) - r^3sin^2(theta) - r^2cos(theta)) dr dtheta

Evaluating the inner integral first, we get:

∫0^(√(33)/4) (4r - 2r^3cos^2(theta) - r^3sin^2(theta) - r^2cos(theta)) dr
= ((2√33)/3)sin(theta) - (17/20)cos(theta) - (2/5)(√33)/3)cos^3(theta) - (1/4)(√33)/3)cos(theta)(sin^2(theta)) - (1/3)sin^3(theta)

Then, integrating with respect to theta, we get:

V = ∫0^2π ((2√33)/3)sin(theta) - (17/20)cos(theta) - (2/5)(√33)/3)cos^3(theta) - (1/4)(√33)/3)cos(theta)(sin^2(theta)) - (1/3)sin^3(theta) dtheta

This integral evaluates to:

V = (5π√33)/3 - (17π)/10

Therefore, the volume of the solid bounded above and below by z=4-x and z=(2*(x^2))+(y^2) is (5π√33)/3 - (17π)/10.

Question: Compute the volume of the solid bounded above and below by the surfaces z=4-x and z= (2*(x^2)) + (y^2) I found the intersection of above 2 surfaces to be the below; (x+ (1/4))^2 + (y/(sqrt(2))^2 = 33/16 How do we proceed from here?

To compute the volume of the solid bounded above and below by the surfaces z=4-x and z=(2*(x^2))+(y^2), we need to integrate the function f(x,y) = 4-x - ((2*x^2) + y^2) over the region R bounded by the intersection curve of the two surfaces.

To compute the volume of the solid bounded above and below by the surfaces z=4-x and z=(2(x^2))+(y^2), we need to integrate the function f(x,y) = 4-x - ((2x^2) + y^2) over the region R bounded by the intersection curve of the two surfaces.