Math, asked by prerananit2019, 4 hours ago

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.​

Answers

Answered by hukam0685
19

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<y<2-x,0<x<1

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.

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