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Find the volume of water of depth $x$ of a conical tank

4

$begingroup$

Find $V(x)$ if $V(x)$ is the volume of water of depth $x$ contained in a conical tank with vertice downwards. The tank is $8$ meters high and its diameter in the highest part is $6$ meters.

Answer:

$V(x)=3pidfracx^364$.

---

I think that a sketch of the situation is:

I tried to use the formula of the conical volume: $V=dfracpi r^2h3$, where $r=dfrac62=3$ and $h=8$, but then $V=dfracpi3^283=24pi$, which 1) does not depend on the depth and 2) does not have the same coefficients of the answer.

What am I doing wrong?

Thanks!

algebra-precalculus volume

share|cite|improve this question

edited Mar 31 at 0:18

Stallmp

21219

asked Mar 30 at 23:31

manoooohmanooooh

6931517

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Your $r=3$ and $h=8$ are for the entire tank. But instead, you need to use the radius and height of the cone that has height $x$.
  $endgroup$
  – Minus One-Twelfth
  Mar 30 at 23:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MinusOne-Twelfth ohh so the volume of that part would be $V(x)=dfracpi(3-x)^2(8-x)3$?
  $endgroup$
  – manooooh
  Mar 30 at 23:37
  

  
  
  
  

add a comment | 

4

$begingroup$

Find $V(x)$ if $V(x)$ is the volume of water of depth $x$ contained in a conical tank with vertice downwards. The tank is $8$ meters high and its diameter in the highest part is $6$ meters.

Answer:

$V(x)=3pidfracx^364$.

---

I think that a sketch of the situation is:

I tried to use the formula of the conical volume: $V=dfracpi r^2h3$, where $r=dfrac62=3$ and $h=8$, but then $V=dfracpi3^283=24pi$, which 1) does not depend on the depth and 2) does not have the same coefficients of the answer.

What am I doing wrong?

Thanks!

algebra-precalculus volume

share|cite|improve this question

edited Mar 31 at 0:18

Stallmp

21219

asked Mar 30 at 23:31

manoooohmanooooh

6931517

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Your $r=3$ and $h=8$ are for the entire tank. But instead, you need to use the radius and height of the cone that has height $x$.
  $endgroup$
  – Minus One-Twelfth
  Mar 30 at 23:35
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MinusOne-Twelfth ohh so the volume of that part would be $V(x)=dfracpi(3-x)^2(8-x)3$?
  $endgroup$
  – manooooh
  Mar 30 at 23:37
  

  
  
  
  

add a comment | 

$begingroup$

Find $V(x)$ if $V(x)$ is the volume of water of depth $x$ contained in a conical tank with vertice downwards. The tank is $8$ meters high and its diameter in the highest part is $6$ meters.

Answer:

$V(x)=3pidfracx^364$.

---

I think that a sketch of the situation is:

I tried to use the formula of the conical volume: $V=dfracpi r^2h3$, where $r=dfrac62=3$ and $h=8$, but then $V=dfracpi3^283=24pi$, which 1) does not depend on the depth and 2) does not have the same coefficients of the answer.

What am I doing wrong?

Thanks!

algebra-precalculus volume

share|cite|improve this question

edited Mar 31 at 0:18

Stallmp

21219

asked Mar 30 at 23:31

manoooohmanooooh

6931517

$endgroup$

share|cite|improve this question

edited Mar 31 at 0:18

Stallmp

21219

asked Mar 30 at 23:31

manoooohmanooooh

6931517

share|cite|improve this question

edited Mar 31 at 0:18

Stallmp

21219

asked Mar 30 at 23:31

manoooohmanooooh

6931517

edited Mar 31 at 0:18

Stallmp

21219

edited Mar 31 at 0:18

Stallmp

21219

asked Mar 30 at 23:31

manoooohmanooooh

6931517

asked Mar 30 at 23:31

manoooohmanooooh

6931517

1 Answer
1

active

oldest

votes

3

$begingroup$

Your issue is that you found the volume of the whole conical tank, not the water.

Imagine taking a vertical cross-section of the tank:

The volume of the water is given by

$$V = frac 1 3 pi r^2 x$$

What is $r$? It can be shown that the triangle formed by the water is similar in the geometric sense to the entire triangle. Then we can set up a proportion:

$$fractextradius of the tanktextheight of the tank = fractextradius of the watertextheight of the water implies frac38 = frac r x implies r = frac 3 8 x$$

Thus,

$$V = frac 1 3 pi left( frac 3 8 x right)^2 x = frac 1 3 cdot pi cdot frac964 cdot x^2 cdot x = frac3pi64x^3$$

matching the answer.

share|cite|improve this answer

edited Mar 30 at 23:56

answered Mar 30 at 23:41

Eevee TrainerEevee Trainer

10.4k31742

$endgroup$

add a comment | 

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3

$begingroup$

Your issue is that you found the volume of the whole conical tank, not the water.

Imagine taking a vertical cross-section of the tank:

The volume of the water is given by

$$V = frac 1 3 pi r^2 x$$

What is $r$? It can be shown that the triangle formed by the water is similar in the geometric sense to the entire triangle. Then we can set up a proportion:

$$fractextradius of the tanktextheight of the tank = fractextradius of the watertextheight of the water implies frac38 = frac r x implies r = frac 3 8 x$$

Thus,

$$V = frac 1 3 pi left( frac 3 8 x right)^2 x = frac 1 3 cdot pi cdot frac964 cdot x^2 cdot x = frac3pi64x^3$$

matching the answer.

share|cite|improve this answer

edited Mar 30 at 23:56

answered Mar 30 at 23:41

Eevee TrainerEevee Trainer

10.4k31742

$endgroup$

add a comment | 

3

$begingroup$

Your issue is that you found the volume of the whole conical tank, not the water.

Imagine taking a vertical cross-section of the tank:

The volume of the water is given by

$$V = frac 1 3 pi r^2 x$$

What is $r$? It can be shown that the triangle formed by the water is similar in the geometric sense to the entire triangle. Then we can set up a proportion:

$$fractextradius of the tanktextheight of the tank = fractextradius of the watertextheight of the water implies frac38 = frac r x implies r = frac 3 8 x$$

Thus,

$$V = frac 1 3 pi left( frac 3 8 x right)^2 x = frac 1 3 cdot pi cdot frac964 cdot x^2 cdot x = frac3pi64x^3$$

matching the answer.

share|cite|improve this answer

edited Mar 30 at 23:56

answered Mar 30 at 23:41

Eevee TrainerEevee Trainer

10.4k31742

$endgroup$

add a comment | 

$begingroup$

Your issue is that you found the volume of the whole conical tank, not the water.

Imagine taking a vertical cross-section of the tank:

The volume of the water is given by

$$V = frac 1 3 pi r^2 x$$

What is $r$? It can be shown that the triangle formed by the water is similar in the geometric sense to the entire triangle. Then we can set up a proportion:

$$fractextradius of the tanktextheight of the tank = fractextradius of the watertextheight of the water implies frac38 = frac r x implies r = frac 3 8 x$$

Thus,

$$V = frac 1 3 pi left( frac 3 8 x right)^2 x = frac 1 3 cdot pi cdot frac964 cdot x^2 cdot x = frac3pi64x^3$$

matching the answer.

share|cite|improve this answer

edited Mar 30 at 23:56

answered Mar 30 at 23:41

Eevee TrainerEevee Trainer

10.4k31742

$endgroup$

share|cite|improve this answer

edited Mar 30 at 23:56

answered Mar 30 at 23:41

Eevee TrainerEevee Trainer

10.4k31742

answered Mar 30 at 23:41

Eevee TrainerEevee Trainer

10.4k31742

answered Mar 30 at 23:41

Eevee TrainerEevee Trainer

10.4k31742