Question

Find the volume of the region bounded above by the paraboloid z= x² + y² and below by the triangle enclosed by the lines y = x, x= 0, and
x + y = 2 in the xy plane.​

Answer 1

Answer:

We need to evaluate the following triple integral:

∫∫∫zdV

The upper and lower limits of z integration are from 0 to 4. To determine the x and y limits we set z=0 and we have

0=4−x2−2y2

which becomes

x2+2y2=4 , an ellipse in the xy plane as illustrated below:

The x limits of integration are from -2 to +2, and the y limits of integration are from −4−x22−−−−√ to 4−x22−−−−√.

Therefore, the volume is calculated as follows

∫2−2∫4−x22√−4−x22√∫40zdzdydx=

∫2−2∫4−x22√−4−x22√z22|40dydx=

∫2−2164−x22−−−−√dx=

162√∫2−24−x2−−−−−√dx=

16π2–√

Answer 2

Given:

The paraboloid $z=x^{2}+y^{2}$ triangle enclosed by the lines $y=x,x=0$.

To find:

The find $x+y=2$ in the $xy$ plane.

Step-by-step explanation:

$V=\int\limits^1_0\int\limits^2_x {x^{2}+y^{2}} \,dydx$

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

$\int\limits^1_0 { \frac{6x^{2}-3x^{3}-x^{3}+6x^{2}-12x+8-4x^{3}}{3} } \, dx$

$=\frac{1}{3}\int\limits^1_0 {12x^{2}-8x^{3}-12x+8} \, dx$

$=\frac{1}{3}[4x^{3}-2x^{4}-6x^{2}+8x]$

$=\frac{1}{3}[4-2-6+8]$

$=\frac{4}{3}$.

Answer:

Therefore, The triangle enclosed by the $x+y=2$ in the $xy$ plane in the $\frac{4}{3}$.