Find the volume of the solid under the plane 3x + 3y − z = 0 and above the region bounded by y = x and y = x^4
To find the volume of the solid under the given plane and above the region bounded by the given curves, we can use a double integral.
Step 1: Determine the limits of integration for x and y.
First, we need to find the limits of integration for x and y within the given region. By setting the two curves equal to each other, we can find the x-values where they intersect:
y = x
y = x^4
Setting these equal to each other:
x = x^4
To solve this equation, we can rearrange it to:
x^4 - x = 0
x(x^3 - 1) = 0
This gives us two solutions: x = 0 and x = 1. Therefore, the limits of integration for x are 0 and 1.
To find the limits of integration for y, we can substitute the values of x into either of the given equations. Let's use y = x:
y = x
So, the limits of integration for y are also 0 and 1.
Step 2: Set up the double integral.
The volume of the solid under the plane can be calculated using the double integral:
V = ∬ R f(x, y) dA
Where R represents the region bounded by the curves, f(x, y) represents the height of the solid at point (x, y), and dA represents the differential area element.
In this case, the height of the solid at any point (x, y) is given by the equation of the plane: 3x + 3y - z = 0.
Rearranging the equation gives:
z = 3x + 3y
Therefore, the double integral becomes:
V = ∬ R (3x + 3y) dA
Step 3: Evaluate the double integral.
Substituting the limits of integration into the integral, we get:
V = ∫[0 to 1] ∫[0 to 1] (3x + 3y) dy dx
Now, we can evaluate each integral separately. Since the integrand is linear, we can do this by integrating term by term:
V = ∫[0 to 1] [3xy + (3y^2)/2] |[0 to 1] dx
Evaluating the expression inside the brackets:
V = ∫[0 to 1] [(3x + 3)/2 - 0] dx
V = ∫[0 to 1] (3x + 3)/2 dx
Integrating each term:
V = [(3x^2)/2 + 3x]/2 |[0 to 1]
Substituting the limits of integration:
V = [(3(1)^2)/2 + 3(1)]/2 - [(3(0)^2)/2 + 3(0)]/2
V = [(3/2) + 3]/2 - [0]/2
V = [(3/2) + 3]/2
V = (3/2 + 6)/2
V = (9/2)/2
Simplifying:
V = 9/4
Therefore, the volume of the solid under the plane 3x + 3y − z = 0 and above the region bounded by y = x and y = x^4 is 9/4 cubic units.
do this like the other. Find the xy-plane boundary of the base, and set the height to
z = 3x+3y