Finding the volume in a cylinder that is intersected by a planeI have a homework question that goes as follows.Let V = { ( x , y , z ) : x 2 + y 2 ≤ 4 and 0 ≤ z ≤ 4 } be a cylinder and let P be the plane through (4, 0, 2), (0, 4, 2), and (-4, -4, 4). Compute the volume of C below the plane P.So I have these points and I set up my a ( x − x 0 ) + b ( y − y 0 ) + c ( z − z 0 ) = 0 where ( x 0 , y 0 , z 0 ) is a point in the plane and ⟨a, b, c⟩ is perpendicular.I then got two vectors that go between two points, namely: P Q =&lt; ( 0 − 4 , ( 4 − 0 ) , ( 2 , − 2 ) &gt;=&lt; − 4 , 4 , 0 &gt; and P R =&lt; ( − 4 − 4 ) , ( − 4 − 0 ) , ( 4 , − 2 ) &gt;=&lt; − 8 , − 4 , 2 &gt;I then get the cross product of them to get &lt; 8 , 8 , 48 &gt; which is the coefficient in my equation of the plane: 8 ( x − 4 ) + 8 ( y − 0 ) + 48 ( z − 2 ) = 0I think that this is the equation of my plane. Is it?

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Finding the volume in a cylinder that is intersected by a planeI have a homework question that goes as follows.Let   V  =  {  (  x  ,  y  ,  z  )  :      x    2    +      y    2    ≤  4   and   0  ≤  z  ≤  4  } be a cylinder and let P be the plane through (4, 0, 2), (0, 4, 2), and (-4, -4, 4). Compute the volume of C below the plane P.So I have these points and I set up my   a  (  x  −      x    0    )  +  b  (  y  −      y    0    )  +  c  (  z  −      z    0    )  =  0 where   (      x    0    ,      y    0    ,      z    0    ) is a point in the plane and ⟨a, b, c⟩ is perpendicular.I then got two vectors that go between two points, namely:  P  Q  =&amp;lt;  (  0  −  4  ,  (  4  −  0  )  ,  (  2  ,  −  2  )  &amp;gt;=&amp;lt;  −  4  ,  4  ,  0  &amp;gt; and   P  R  =&amp;lt;  (  −  4  −  4  )  ,  (  −  4  −  0  )  ,  (  4  ,  −  2  )  &amp;gt;=&amp;lt;  −  8  ,  −  4  ,  2  &amp;gt;I then get the cross product of them to get   &amp;lt;  8  ,  8  ,  48  &amp;gt; which is the coefficient in my equation of the plane:  8  (  x  −  4  )  +  8  (  y  −  0  )  +  48  (  z  −  2  )  =  0I think that this is the equation of my plane. Is it?

wlanauee · Accepted Answer

Step 1First of all, you made a mistake in your second vector when computing the direction of the plane:  P  R  =  (  −  8  ,  −  4  ,  2  )and your cross product doesn&#039;t look correct even with your vector PR.Now the volume can be set up like this:      ∬    S    z  d  x  d  ywhere S is the projection of the volume to the xy plane and z is the upper limit of z which is the plane represented by z.Step 2In fact, you will need to check whether the plane is below   z  =  4 in the confined region. After some trick you will see the plane is really below   z  =  4. That&#039;s why the   z  =  4 would not be in your integral any more.The triple integral is   ∫  ∫      ∫    0    z    d  z  d  x  d  ywhere z is exactly the plane. That is why I wrote the double integral.

Ignacio Riggs · Accepted Answer

Step 1If we let             a      →        =  ⟨  −  4  ,  4  ,  0  ⟩ and             b      →        =  ⟨  −  8  ,  −  4  ,  2  ⟩, then             a      →        ×            b      →        =  ⟨  8  ,  8  ,  48  ⟩.Then we can take             n      →        =  ⟨  1  ,  1  ,  6  ⟩ as a normal vector to the plane, so substituting in the coordinates of one of the points gives   x  +  y  +  6  z  =  16.Step 2Now you can use polar coordinates or rectangular coordinates to find the volume,since the base of the solid is the region in the xy-plane bounded by the circle       x    2    +      y    2    =  4,and the height of the solid is given by   z  =      1    6    (  16  −  x  −  y  )

Let V={(x,y,z):x^2+y^2 leq 4 and 0 leq z leq 4} be a cylinder and let P be the plane through (4, 0, 2),(0, 4, 2), and (-4, -4, 4). Compute the volume of C below the plane P.

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