![\bold{\huge{\underline{ Solution }}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/jdi2w7914cic76zpb2xuxp7e51pz44d9g8.png)
Given :-
- The length of the part is 16 centimeters
- The part of new engine is composed of 1 cone, 1 cylinder and 1 hemisphere
- The width of the engine is 6 cm
- The height of the cone is 5 cm
To Find :-
- We have to find the total volume of the part
Let's Begin :-
Let divide the part of engine into three parts as it is composed of 3 different figures.
We know that,
Volume of cone
![\bold{=}{\bold{( 1)/(3)}}{\bold{{\pi}r^(2)h}}](https://img.qammunity.org/2023/formulas/mathematics/college/mhyj2xcezoyrzj8hzl3gl3o00l7v3ukklg.png)
Here, we have,
- The height of the cone is 5 cm
- The diameter of the cone is 6 cm
- Therefore,
- Radius of the cone = 3 cm
Subsitute the required values,
Volume of the first part that is cone
![\sf{=}{\sf{( 1)/(3)}}{\sf{ {*}3.14{*}(3)^(2){*}5}}](https://img.qammunity.org/2023/formulas/mathematics/college/ao6c13p9ngtoyvjwj1dpv5ih4urs1vdqmq.png)
![\sf{=}{\sf{( 1)/(3)}}{\sf{ {*}3.14{*}9{*}5}}](https://img.qammunity.org/2023/formulas/mathematics/college/6yms6e703pv3o5fjcdne3prs227wv52c71.png)
![\sf{ = 3.14 {*} 3 {*} 5 }](https://img.qammunity.org/2023/formulas/mathematics/college/g8054axxemr54f00zb2yvalkygb5s658on.png)
![\sf{ = 3.14 {*} 15 }](https://img.qammunity.org/2023/formulas/mathematics/college/qgm37jfadf64y1svc7pfrbqp0uat9jha7l.png)
![\sf{ = 3.14 {*} 15 }](https://img.qammunity.org/2023/formulas/mathematics/college/qgm37jfadf64y1svc7pfrbqp0uat9jha7l.png)
![\bold{ = 47.1 cm^(3) }](https://img.qammunity.org/2023/formulas/mathematics/college/q9r7hfwdkgdslrjj1gv6h3gcqy4iaytmi5.png)
Thus, The volume of cone is 47.1 cm³ .
For second part
- Second part is composed of cylinder
We know that,
The volume of cylinder
![\bold{ = {\pi}r^(2)h }](https://img.qammunity.org/2023/formulas/mathematics/college/29uesvkbrzjlp9l48yh22lt3180a0mg6v3.png)
Here,
- The diameter of the cylinder is 6 cm
- So, Radius = 3 cm
- The length of the cylinder = 16 - (Length of cone + Length of hemisphere)
- Length = 16 - 11 = 5 cm
Subsitute the required values in the above formula,
Volume of the second part that is cylinder
![\sf{ = 3.14{*} (3)^(2){*} 5}](https://img.qammunity.org/2023/formulas/mathematics/college/lajuiduhne23rfgmgl9gdhmaqp4zbwdq1v.png)
![\sf{ = 3.14{*} 9 {*} 5}](https://img.qammunity.org/2023/formulas/mathematics/college/qtn8c72pdpqs9nl2jov9uqe5qg0y5knh99.png)
![\sf{ = 3.14{*} 45 }](https://img.qammunity.org/2023/formulas/mathematics/college/cqwmka0bafmqecmyzokaumd76r78h72mxb.png)
![\bold{ = 141.3 cm^(3)}](https://img.qammunity.org/2023/formulas/mathematics/college/57w1zlqh5wq6c4cmkfdduubl2irmdhxpkz.png)
Thus, The volume of the cylinder is 141.3 cm³
For third part
- Third part is composed of hemisphere
We know that,
Volume of hemisphere
![\bold{=}{\bold{( 2)/(3)}}{\sf{ {\pi}r}}](https://img.qammunity.org/2023/formulas/mathematics/college/kwir2yrermtvq9pf8pqpsqol36oduzpyz5.png)
Here,
- The diameter of the hemisphere is 6 cm
- So, Radius = 3cm
Subsitute the required values,
Volume of third part that is hemisphere
![\sf{=}{\sf{( 2)/(3)}}{\sf{{*} 3.14 {*}3}}](https://img.qammunity.org/2023/formulas/mathematics/college/ijpswqypa2z5mbd6ch39hxbpktbq1qayc6.png)
![\sf{ = 2 {*} 3.14 }](https://img.qammunity.org/2023/formulas/mathematics/college/sqhq8w9g293vpalfa4a5quxf4twrhv5d1j.png)
![\bold{ = 6.28 cm^(3) }](https://img.qammunity.org/2023/formulas/mathematics/college/sfmcewvwnowv8a79g7xurqxiq8fekhkqn6.png)
Thus, The volume of the hemisphere is 6.28 cm³
Therefore,
The total volume of the part
= Volume of cone + Volume of cylinder + Volume of hemisphere
![\sf{ = 47.1 + 141.3 + 6.28 }](https://img.qammunity.org/2023/formulas/mathematics/college/pwn0kvz2fujd1eq0skccnku2tb800ubhay.png)
![\sf{ = 188.4 + 6.28 }](https://img.qammunity.org/2023/formulas/mathematics/college/2c0ak4o8ogmbctii3w01g14g92j8g0hmy2.png)
![\sf{ = 194.68 cm^(3) }](https://img.qammunity.org/2023/formulas/mathematics/college/6oymr6fvggynovn1rculmthkc38nrmt0nx.png)
![\bold{ = 194.7 cm^(3) }](https://img.qammunity.org/2023/formulas/mathematics/college/ibo10fjcop7ewb8f61kkl00k81ybdv3eqj.png)
Hence, The total volume of the part is 194.7 cm³ .