Find the volume under the paraboloid z = x2 + y2 above the triangle en

1.	Find the volume under the paraboloid z = x2 + y2 above the triangle enclosed by the lines y = x, x = 0 and x + y = 2 in the xy-plane.
Answer» Step-by-step-Explanation: Given: Paraboloid z = x² + y² To FIND:Find the volume under the paraboloid z = x² + y² above the triangle enclosed by the lines y = x, x = 0 and x + y = 2 in the xy-plane. Solution: $V=\int\int_R x^2+y^2 dA\\$ First find the limits of double integral using the given conditions y = x, x = 0 and x + y = 2 in the xy-plane. Region R: x Now,PUT the limits and INTEGRATE to find the volume of Paraboloid $\bold{\red{V=\int_0^1 \int_x^{2-x} x^2+y^2 dy dx}}\\$ Integrate with respect to y $V=\int_0^1\left[x^2 y+\frac{y^3}{3}\right]_x^{2-x} dx\\$ Put the limits $V=\int_0^1\left[x^2 (2-x)+\frac{(2-x)^3}{3}-x^2 (x)-\frac{(x)^3}{3}\right] dx\\$ $V=\int_0^1\left[ (2-x)(x^2+\frac{(2-x)^2}{3})-x^2 (x)-\frac{(x)^3}{3}\right] dx\\$ $V=\int_0^1\left[ (2-x).\frac{ 3{x}^{2} + {x}^{2} - 4x + 4 }{3}-\frac{3 {x}^{3} + x^3}{3}\right] dx\\ \\ V=\int_0^1\left[\frac{8{x}^{2}-8x+8- 4x^{3} + 4x^2-4x}{3}-\frac{4 {x}^{3}}{3}\right] dx\\ \\ V= \frac{1}{3} \int_0^1 - 8 {x}^{3} + 12 {x}^{2} - 12x + 8 \: dx$ $V= \frac{1}{3}\left[\frac{- 8 {x}^{4}}{4} + \frac{12 {x}^{3}}{3} -\frac{ 12x^2 }{2} + 8x \right]_0^1\\$ $V= \frac{1}{3}\left[- 2 {x}^{4} + 4 {x}^{3} -6x^2 + 8x \right]_0^1\\ \\V= \frac{1}{3}\left[- 2 {(1)}^{4} + 4 {(1)}^{3} -6(1)^2 + 8(1) + 2 {(0)}^{4} - 4 {(0)}^{3} + 6(0)^2 - 8(0) \right]\\ \\ \\V= \frac{1}{3}\left[- 2+ 4 -6 + 8\right] \\ \\ \bold{\green{V = \frac{4}{3} }} \\$ Final answer: Volume of paraboloid z = x² + y² above the triangle enclosed by the lines y = x, x = 0 and x + y = 2 in the xy-plane is 4/3 cube units. Hope it HELPS you.

Find the volume under the paraboloid z = x2 + y2 above the triangle enclosed by the lines y = x, x = 0 and x + y = 2 in the xy-plane.

Answer»

Step-by-step-Explanation:

Given: Paraboloid z = x² + y²

To FIND:Find the volume under the paraboloid z = x² + y² above the triangle enclosed by the lines y = x, x = 0 and x + y = 2 in the xy-plane.

Solution:

$V=\int\int_R x^2+y^2 dA\\$

First find the limits of double integral using the given conditions y = x, x = 0 and x + y = 2 in the xy-plane.

Region R: x

Now,PUT the limits and INTEGRATE to find the volume of Paraboloid

$\bold{\red{V=\int_0^1 \int_x^{2-x} x^2+y^2 dy dx}}\\$

Integrate with respect to y

$V=\int_0^1\left[x^2 y+\frac{y^3}{3}\right]_x^{2-x} dx\\$

Put the limits

$V=\int_0^1\left[x^2 (2-x)+\frac{(2-x)^3}{3}-x^2 (x)-\frac{(x)^3}{3}\right] dx\\$

$V=\int_0^1\left[ (2-x)(x^2+\frac{(2-x)^2}{3})-x^2 (x)-\frac{(x)^3}{3}\right] dx\\$

$V=\int_0^1\left[ (2-x).\frac{ 3{x}^{2} + {x}^{2} - 4x + 4 }{3}-\frac{3 {x}^{3} + x^3}{3}\right] dx\\ \\ V=\int_0^1\left[\frac{8{x}^{2}-8x+8- 4x^{3} + 4x^2-4x}{3}-\frac{4 {x}^{3}}{3}\right] dx\\ \\ V= \frac{1}{3} \int_0^1 - 8 {x}^{3} + 12 {x}^{2} - 12x + 8 \: dx$

$V= \frac{1}{3}\left[\frac{- 8 {x}^{4}}{4} + \frac{12 {x}^{3}}{3} -\frac{ 12x^2 }{2} + 8x \right]_0^1\\$

$V= \frac{1}{3}\left[- 2 {x}^{4} + 4 {x}^{3} -6x^2 + 8x \right]_0^1\\ \\V= \frac{1}{3}\left[- 2 {(1)}^{4} + 4 {(1)}^{3} -6(1)^2 + 8(1) + 2 {(0)}^{4} - 4 {(0)}^{3} + 6(0)^2 - 8(0) \right]\\ \\ \\V= \frac{1}{3}\left[- 2+ 4 -6 + 8\right] \\ \\ \bold{\green{V = \frac{4}{3} }} \\$

Final answer:

Volume of paraboloid z = x² + y² above the triangle enclosed by the lines y = x, x = 0 and x + y = 2 in the xy-plane is 4/3 cube units.

Hope it HELPS you.