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Answer: [tex]\int\limits^a_E {\int\limits^a_E {\int\limits^a_E {xy} } \, dV[/tex] = 1087.5
Step-by-step explanation: To evaluate the triple integral, first an equation of a plane is needed, since the tetrahedon is a geometric form that occupies a 3 dimensional plane. The region of the integral is in the attachment.
An equation of a plane is found with a point and a normal vector. Normal vector is a perpendicular vector on the plane.
Given the points, determine the vectors:
P = (5,0,0); Q = (0,9,0); R = (0,0,4)
vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)
vector QR = (0,9,0) - (0,0,4) = (0,9,-4)
Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:
n = PQ × QR = [tex]\left[\begin{array}{ccc}i&j&k\\5&-9&0\\0&9&-4\end{array}\right][/tex][tex]\left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right][/tex]
n = 36i + 0j + 45k - (0k + 0i - 20j)
n = 36i + 20j + 45k
Equation of a plane is generally given by:
[tex]a(x-x_{0}) + b(y-y_{0}) + c(z-z_{0}) = 0[/tex]
Then, replacing with point P and normal vector n:
[tex]36(x-5) + 20(y-0) + 45(z-0) = 0[/tex]
The equation is: 36x + 20y + 45z - 180 = 0
Second, in evaluating the triple integral, set limits:
In terms of z:
[tex]z = \frac{180-36x-20y}{45}[/tex]
When z = 0:
[tex]y = 9 + \frac{-9x}{5}[/tex]
When z=0 and y=0:
x = 5
Then, triple integral is:
[tex]\int\limits^5_0 {\int\limits {\int\ {xy} \, dz } \, dy } \, dx[/tex]
Calculating:
[tex]\int\limits^5_0 {\int\limits {\int\ {xyz} \, dy } \, dx[/tex]
[tex]\int\limits^5_0 {\int\limits {\int\ {xy(\frac{180-36x-20y}{45} - 0 )} \, dy } \, dx[/tex]
[tex]\frac{1}{45} \int\limits^5_0 {\int\ {180xy-36x^{2}y-20xy^{2}} \, dy } \, dx[/tex]
[tex]\frac{1}{45} \int\limits^5_0 {90xy^{2}-18x^{2}y^{2}-\frac{20}{3} xy^{3} } \, dx[/tex]
[tex]\frac{1}{45} \int\limits^5_0 {2430x-1458x^{2}+\frac{94770}{125} x^{3}-\frac{23490}{375}x^{4} } \, dx[/tex]
[tex]\frac{1}{45} [30375-60750+118462.5-39150][/tex]
[tex]\int\limits^5_0 {\int\limits {\int\ {xyz} \, dy } \, dx[/tex] = 1087.5
The volume of the tetrahedon is 1087.5 cubic units.
The tripple integration will be [tex]\int\limits^a_E \int\limits^a_E \int\limits^a_E {xy} \, dV[/tex] = 1087.5
What is triple integration?
The triple integration is used to identify the volumes of the objects or for analyzing three dimension of the object.
To evaluate the triple integral, first an equation of a plane is needed, since the tetrahedron is a geometric form that occupies a 3 dimensional plane. The region of the integral is in the attachment.
An equation of a plane is found with a point and a normal vector. Normal vector is a perpendicular vector on the plane.
Given the points, determine the vectors:
P = (5,0,0); Q = (0,9,0); R = (0,0,4)
vector PQ = (5,0,0) - (0,9,0) = (5,-9,0)
vector QR = (0,9,0) - (0,0,4) = (0,9,-4)
Knowing that cross product of two vectors will be perpendicular to these vectors, you can use the cross product as normal vector:
n = PQ × QR = [tex]\left[\begin{array}{ccc}i&j&k\\5&-9&-0\\0&9&-4\end{array}\right] \left[\begin{array}{ccc}i&j\\5&-9\\0&9\end{array}\right][/tex]
n = 36i + 0j + 45k - (0k + 0i - 20j)
n = 36i + 20j + 45k
Equation of a plane is generally given by:
[tex]a(x-x_o)+b(y-y_o)+c(z-z_o)=0[/tex]
Then, replacing with point P and normal vector n:
[tex]36(x-5)+20(y-0)+45(z-0)=0[/tex]
The equation is: 36x + 20y + 45z - 180 = 0
Second, in evaluating the triple integral, set limits:
In terms of z:
[tex]z=\dfrac{180-36x-20y}{45}[/tex]
When z = 0:
[tex]y=9+\dfrac{-9x}{5}[/tex]
When z=0 and y=0:
x = 5
Then, triple integral is:
[tex]\int\limits^5_0 \int\int xy\ dzdydx[/tex]
Calculating:
[tex]\int\limits^5_0 \int\int xy\ dzdydx[/tex]
[tex]\int\limits^5_0 \int\int xy\ (\dfrac{180-36x-20y}{45}-0)dydx[/tex]
[tex]\dfrac{1}{45}\int\limits^5_0 \int \ 180xy-36x^2y-20xy^2dydx[/tex]
[tex]\dfrac{1}{45}\int\limits^5_0 \int \ 90xy^2-18x^2y^2-\dfrac{20}{3}xy^3dydx[/tex]
[tex]\dfrac{1}{45}\int\limits^5_0 \int \ 2430x-1458x^2+\dfrac{94770}{125}x^3-\dfrac{23490}{375}x^4dx[/tex]
[tex]\dfrac{1}{45} [30375-60750+118462.5-39150][/tex]
[tex]\int\limits^5_0 \int\int xy\ dzdydx=1087.5[/tex]
The volume of the tetrahedron is 1087.5 cubic units.
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