Mathematics, 07.04.2020 23:09 kaydenceskyem

Use a double integral to find the volume of the solid in the first octant bounded by the plane determined by the three non-colinear points P(3, 0, 0), Q(0, 1, 0),R(0, 0, 2). Draw a picture of the volume you are finding.

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Answer from: joejoefofana

2.5 units^3

Step-by-step explanation:

Given:-

- The solid is bounded by a plane defined by the following points:

P(3, 0, 0), Q(0, 1, 0),R(0, 0, 2)

Find:-

Use a double integral to find the volume of the solid in the first octant bounded by the plane

Solution:-

- Determine the equation of the plane. Compute two direction vectors d1 and d2 that lie on the plane:

d1 = P - Q

d1 = (3, 0, 0) - (0, 1, 0) = (3,-1,0)

d2 = P - R

d2 = (3, 0, 0) - (0, 0, 2) = (3,0,-2)

- Find the a vector "normal" - n to the plane by cross product formulation of direction vectors (d1 and d2) that lie on the plane:

$n = d1xd2 = \left[\begin{array}{c}3&-1&0\end{array}\right] x \left[\begin{array}{c}3&0&-2\end{array}\right] = \left[\begin{array}{ccc}3&-1&0\\3&0&2\end{array}\right] = \left[\begin{array}{c}-2&-6&3\end{array}\right]$

- The equation of plane is:

n.(x,y,z) = n.P

-2x -6y + 3z = -6

- The function of one variable would be:

z = (2/3)x + 2y - 2

- The double integration formulation would be:

$\int\limits^a_b \int\limits^c_d f(z) dy.dx\\\\\int\limits^a_b \int\limits^c_d (\frac{2x}{3} + 2y - 3) dy.dx\\\\\int\limits^a_b (\frac{2xy}{3} + y^2 - 3y) |_c^d.dx$

- Where the limits (c and d) are defined by planar (x-y) projection of plane (n) :

y = d = -(1/3)x + 1

c = 0

- Evaluate the limits:

$\int\limits^a_b (\frac{-2x^2 + 6x}{9} + \frac{x^2}{9} -\frac{2x}{3} +1 + x - 3) .dx\\\\\int\limits^a_b (\frac{-x^2 }{9} + x - 2) .dx\\\\(\frac{-x^3 }{27} + \frac{x^2}{2} - 2x)|^3_0\\\\(\frac{-27 }{27} + \frac{9}{2} - 6) = 1 - 4.5 + 6 = 2.5 unit^3$

Answer from: Quest

you didn't give me the choices, so i will just give you the graph so you can look at your choices and choose the right one.

Answer from: Quest

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in $\delta bea$ and $\delta dea$ ;

$ba=da$ , as these are radii of our given circle.

$ea=ea$

therefore, $\delta bea\cong \delta dea$ by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

$3x-24=x+10$