"1) Find the area of the part of..." solved and explained

College
Calculus and Analysis
Integral Calculus
Parametric equations, polar coordinates, and vector-valued functions

Emeli Hagan

Answered question

2021-06-08

1) Find the area of the part of the plane
$4 x + 3 y + z = 12$
that lies in the first octant.
2) Use polar coordinates to find the volume of the given solid.
Bounded by the paraboloid $z = 5 + 2 x^{2} + 2 y^{2}$ and the plane z = 11 in the first octant

Answer & Explanation

aprovard

Skilled2021-06-09Added 94 answers

I have resolved your question. The answer is below:

Jeffrey Jordon

Expert2021-09-30Added 2605 answers

1) $4 x + 3 y + 3 = 12$

$\vec{R} (x, y) =< x, y, 12 - 4 x - 3 y >$

$\Rightarrow | \vec{R_{x}} \times \vec{R_{y}} | = \sqrt{(f_{x})^{2} + (f_{y})^{2} + 1} = \sqrt{(- 4)^{2} + (- 3)^{2} + 1}$

$= \sqrt{16 + 9 + 1} = \sqrt{26}$

$\int_{S} \int d s = \int \int_{D} \sqrt{26} d A$

$(x, y, 3)$ axis $= (3, 0, 0) (0, 4, 0) (0, 0, 12)$

So, the domain D in triangle is bounded by $(0, 0) (3, 0) (0, 4)$

$\int \int_{D} \sqrt{26} d A = \sqrt{26} \times \frac{1}{2} \times 3 \times 4^{2} = 6 \sqrt{26}$

$z = 5 + 2 x^{2} + 2 y^{2}$

$\Rightarrow 11 = 5 + 2 x^{2} + 2 y^{2}$

$\Rightarrow 6 = 2 x^{2} + 2 y^{2} \Rightarrow x^{2} + y^{2} = 3 | x^{2} + y^{2} = R^{2}$

Volume $= \int_{0}^{\frac{π}{2}} \int_{0}^{\sqrt{3}} (5 + 2 x^{2} + 2 y^{2}) d A$

$= \int_{0}^{\frac{π}{2}} \int_{0}^{\sqrt{3}} (5 + 2 R) R d R d Q = \int_{0}^{\frac{π}{2}} \int_{0}^{\sqrt{3}} 3 R + 2 R^{2} d r d Q$

$= \int_{0}^{\frac{π}{2}} [\frac{5 R^{2}}{8} + \frac{2 R^{3}}{3}]_{0}^{\sqrt{3}} d Q = \int_{0}^{\frac{π}{2}} \frac{15}{2} + \frac{6}{\sqrt{3}} d Q$

$\Rightarrow \frac{15 π}{4} + \sqrt{3} π = π [\frac{15}{4} + \sqrt{3}]$ Answer

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Integral Calculus

Didn't find what you were looking for?