Respuesta :
Answer:
Step-by-step explanation:
Given
Paraboloid [tex]z=x^2+y^2[/tex] and Plane [tex]x+y=1[/tex] are bounded to get the area
In first octant
value of x varies from [tex]0\leq x\leq 1[/tex]
value of y varies from [tex]0\leq y\leq 1[/tex]
Volume [tex]V=\int_{0}^{1}\int_{0}^{1-x}\left ( x^2+y^2\right )dydx[/tex]
[tex]V=\int_{0}^{1}\left [ x^2y+\frac{y^3}{3}\right ]_{0}^{1-x}[/tex]
[tex]V=\int_{0}^{1}\left [ x^2(1-x)+\frac{(1-x)^3}{3}\right ]dx[/tex]
[tex]V=\int_{0}^{1}\left ( \frac{-4x^3}{3}+2x^2-x+\frac{1}{3}\right )dx[/tex]
[tex]V=\frac{1}{6}[/tex]
The volume of the region bounded by the paraboloid z=x2+y2 and the plane x+y=1 in the first octant is 1/6
The paraboloid and the plane are given as:
[tex]z=x^2+y^2[/tex]
[tex]x+y = 1[/tex]
Make y the subject
[tex]y = 1 - x[/tex]
The plane is in the first octant.
So, the volume is represented as:
[tex]V = \int\limits^1_0 { \int\limits^y_0 {z^2} \, dy } \, dx[/tex]
The integral becomes
[tex]V = \int\limits^1_0 { \int\limits^{1-x}_0 {x^2 + y^2} \, dy } \, dx[/tex]
Integrate with respect to y
[tex]V = \int\limits^1_0 [x^2y + \frac 13y^3]\limits^{1-x}_0 } \, dx[/tex]
Expand
[tex]V = \int\limits^1_0 [x^2(1-x) + \frac 13(1-x)^3]} \, dx[/tex]
Expand
[tex]V = \int\limits^1_0 [x^2-x^3 + \frac 13(1 -x + x^2 - x^3)]} \, dx[/tex]
Expand
[tex]V = \int\limits^1_0 [x^2-x^3 + \frac 13 -x + x^2 - \frac {x^3}3]} \, dx[/tex]
Evaluate the like terms
[tex]V = \int\limits^1_0 [ \frac 13 -x + 2x^2 - \frac {4x^3}3]} \, dx[/tex]
Integrate
[tex]V = [\frac 13x -\frac 12x^2 + \frac 23x^3 - \frac 13x^4]\limits^1_0[/tex]
Expand
[tex]V = \frac 13(1) -\frac 12(1)^2 + \frac 23(1)^3 - \frac 13(1)[/tex]
[tex]V = -\frac 12(1)^2 + \frac 23(1)^3[/tex]
Evaluate the exponents
[tex]V = -\frac 12 + \frac 23[/tex]
Add the fractions
[tex]V = \frac{-3 + 4}{6}[/tex]
[tex]V = \frac{1}{6}[/tex]
Hence, the volume of the region is 1/6
Read more about volumes at:
https://brainly.com/question/8994737
