Active forum topics
- Hydraulics: Rotating Vessel
- Inverse Trigo
- General Solution of y′=xlnx
- engineering economics: construct the cash flow diagram
- Eliminate the Arbitrary Constants
- Law of cosines
- Maxima and minima (trapezoidal gutter)
- Special products and factoring
- Integration of 4x^2/csc^3x√sinxcosx dx
- application of minima and maxima
New forum topics
Recent comments
- Determine the least depth…1 week 3 days ago
- Solve mo ang h manually…3 weeks ago
- Paano kinuha yung height na…3 weeks ago
- It's the unit conversion…1 month ago
- Refer to the figure below…3 weeks 6 days ago
- Yes.4 months 2 weeks ago
- Sir what if we want to find…4 months 2 weeks ago
- Hello po! Question lang po…5 months 1 week ago
- 400000=120[14π(D2−10000)]
(…6 months 1 week ago - Use integration by parts for…7 months 1 week ago
To get the height of this
To get the height of this weird cylinder, recall the formula:
V=BH
where V is the volume of the figure, B is the area of the base, and H is the height of the figure.
With that in mind, we plug this given to the formula above:
V=BH 350 in3=(12r2θ)(4r)
350 in3=(12(r2)(25o(π180o)))(4r) 350 in3=(12(r2)(5π36))(4r) 350 in3=5π18r3 1260π=r3 r=7.3 inches
Note that the area of a sector of circle is A=12r2θ, where r is the radius of the sector and θ is the central angle of the sector.
Since the problem asks for the height of this weird cylinder, which is four times longer than the radius of the sector of circle, so:
H=4r H=4(7.3 in) H=29.2 in
Therefore, the height of this weird cylinder is H=29.2 in.
Alternate solutions are encouraged...