Question

Find the volume of the region bounded by the paraboloid z=x2+y2 and the plane x+y=1 in the first octant.

Answer 1

Answer:

Step-by-step explanation:

Given

Paraboloid $z=x^2+y^2$ and Plane $x+y=1$ are bounded to get the area

In first octant

value of x varies from $0\leq x\leq 1$

value of y varies from $0\leq y\leq 1$

Volume $V=\int_{0}^{1}\int_{0}^{1-x}\left ( x^2+y^2\right )dydx$

$V=\int_{0}^{1}\left [ x^2y+\frac{y^3}{3}\right ]_{0}^{1-x}$

$V=\int_{0}^{1}\left [ x^2(1-x)+\frac{(1-x)^3}{3}\right ]dx$

$V=\int_{0}^{1}\left ( \frac{-4x^3}{3}+2x^2-x+\frac{1}{3}\right )dx$

$V=\frac{1}{6}$

Answer 2

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:

$z=x^2+y^2$

$x+y = 1$

Make y the subject

$y = 1 - x$

The plane is in the first octant.

So, the volume is represented as:

$V = \int\limits^1_0 { \int\limits^y_0 {z^2} \, dy } \, dx$

The integral becomes

$V = \int\limits^1_0 { \int\limits^{1-x}_0 {x^2 + y^2} \, dy } \, dx$

Integrate with respect to y

$V = \int\limits^1_0 [x^2y + \frac 13y^3]\limits^{1-x}_0 } \, dx$

Expand

$V = \int\limits^1_0 [x^2(1-x) + \frac 13(1-x)^3]} \, dx$

Expand

$V = \int\limits^1_0 [x^2-x^3 + \frac 13(1 -x + x^2 - x^3)]} \, dx$

Expand

$V = \int\limits^1_0 [x^2-x^3 + \frac 13 -x + x^2 - \frac {x^3}3]} \, dx$

Evaluate the like terms

$V = \int\limits^1_0 [ \frac 13 -x + 2x^2 - \frac {4x^3}3]} \, dx$

Integrate

$V = [\frac 13x -\frac 12x^2 + \frac 23x^3 - \frac 13x^4]\limits^1_0$

Expand

$V = \frac 13(1) -\frac 12(1)^2 + \frac 23(1)^3 - \frac 13(1)$

$V = -\frac 12(1)^2 + \frac 23(1)^3$

Evaluate the exponents

$V = -\frac 12 + \frac 23$

Add the fractions

$V = \frac{-3 + 4}{6}$

$V = \frac{1}{6}$

Hence, the volume of the region is 1/6

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