Examkin

Soru

Zorluk: Çok zorVolume and Surface Area of Solids

A solid metal right circular cylinder has a height of hh and a base radius of rr, where h>2rh > 2r. From one flat end of the cylinder, a hemisphere of radius rr is carved out. From the other flat end, a right circular cone of base radius rr and height h−2rh - 2r is carved out. If the volume of the remaining solid is equal to the volume of a sphere with radius RR, which of the following equations represents RR in terms of rr and hh?

  1. R=r2h23R = \sqrt[3]{\frac{r^2 h}{2}}Cevap
  2. B
    R=r2h2R = \sqrt{\frac{r^2 h}{2}}
  3. C
    R=r2h3R = \sqrt[3]{r^2 h}
  4. D
    R=r2h2R = \frac{r^2 h}{2}

Cevap

The correct equation is R=r2h23R = \sqrt[3]{\frac{r^2 h}{2}}
The volume of the cylinder is πr2h\pi r^2 h. Subtracting the hemisphere's volume (23πr3\frac{2}{3}\pi r^3) and the cone's volume (13πr2(h−2r)=13πr2h−23πr3\frac{1}{3}\pi r^2(h - 2r) = \frac{1}{3}\pi r^2 h - \frac{2}{3}\pi r^3) leaves a remaining volume of 23πr2h\frac{2}{3}\pi r^2 h. Setting this equal to the volume of a sphere of radius RR (43πR3\frac{4}{3}\pi R^3) gives 43πR3=23πr2h\frac{4}{3}\pi R^3 = \frac{2}{3}\pi r^2 h. Simplifying this equation yields R3=r2h2R^3 = \frac{r^2 h}{2}, and taking the cube root gives the correct relation.

Adım Adım Çözüm

1
Write the formula for the volume of the original cylinder.
Vcylinder=πr2hV_{\text{cylinder}} = \pi r^2 h
The volume of a cylinder with radius rr and height hh is given by the formula πr2h\pi r^2 h.
2
Write the volume formulas for the carved-out shapes.
Vhemisphere=23πr3V_{\text{hemisphere}} = \frac{2}{3}\pi r^3 and Vcone=13πr2(h−2r)V_{\text{cone}} = \frac{1}{3}\pi r^2 (h - 2r)
A hemisphere is half a sphere, so its volume is 23πr3\frac{2}{3}\pi r^3. The volume of a cone with radius rr and height h−2rh - 2r is 13πr2(h−2r)\frac{1}{3}\pi r^2 (h - 2r).
3
Subtract the volumes of the carved-out shapes from the cylinder's volume.
Vremaining=πr2h−23πr3−13πr2(h−2r)=23πr2hV_{\text{remaining}} = \pi r^2 h - \frac{2}{3}\pi r^3 - \frac{1}{3}\pi r^2 (h - 2r) = \frac{2}{3}\pi r^2 h
Expanding the cone's volume gives 13πr2h−23πr3\frac{1}{3}\pi r^2 h - \frac{2}{3}\pi r^3. Subtracting this and the hemisphere's volume from the cylinder's volume simplifies to 23πr2h\frac{2}{3}\pi r^2 h because the πr3\pi r^3 terms cancel out.
4
Equate the remaining volume to the volume of a sphere with radius RR and solve for RR.
43πR3=23πr2h  ⟹  R3=r2h2  ⟹  R=r2h23\frac{4}{3}\pi R^3 = \frac{2}{3}\pi r^2 h \implies R^3 = \frac{r^2 h}{2} \implies R = \sqrt[3]{\frac{r^2 h}{2}}
Equating the remaining volume to the volume of a sphere of radius RR and dividing both sides by 23π\frac{2}{3}\pi gives 2R3=r2h2 R^3 = r^2 h. Dividing by 2 and taking the cube root isolates RR.

Anahtar Kavram

Calculating and comparing volumes of combined geometric solids and algebraically isolating variables.
Tahmini Süre:3m 0s
Bu soruyu puanla

Konular

Volume and Surface Area of Solids

Bu konudan daha fazla soru