Let's solve the problem step-by-step.
1. Understanding the problem:
- We have a cylinder with:
- Radius \( r = 8 \) units
- Height \( h = 15 \) units
- Inside this cylinder, there is an oblique cone with:
- The same radius of the base \( r = 8 \) units
- Slant height \( l = 17 \) units
2. Volume of the cylinder:
- The formula for the volume of a cylinder is:
[tex]\[
V_{\text{cylinder}} = \pi r^2 h
\][/tex]
- Substituting the values:
[tex]\[
V_{\text{cylinder}} = \pi (8)^2 (15) = \pi \times 64 \times 15 = 960 \pi \quad \text{cubic units}
\][/tex]
3. Determining the height of the cone:
- We use the Pythagorean theorem for the triangle formed by the radius, height, and slant height of the cone:
[tex]\[
l^2 = r^2 + h_{\text{cone}}^2
\][/tex]
[tex]\[
17^2 = 8^2 + h_{\text{cone}}^2
\][/tex]
[tex]\[
289 = 64 + h_{\text{cone}}^2
\][/tex]
[tex]\[
h_{\text{cone}}^2 = 225
\][/tex]
[tex]\[
h_{\text{cone}} = \sqrt{225} = 15 \quad \text{units}
\][/tex]
4. Volume of the cone:
- The formula for the volume of a cone is:
[tex]\[
V_{\text{cone}} = \frac{1}{3} \pi r^2 h_{\text{cone}}
\][/tex]
- Substituting the values:
[tex]\[
V_{\text{cone}} = \frac{1}{3} \pi (8)^2 (15) = \frac{1}{3} \pi \times 64 \times 15 = \frac{960 \pi}{3} = 320 \pi \quad \text{cubic units}
\][/tex]
5. Unfilled volume inside the cylinder:
- The unfilled volume is the volume of the cylinder minus the volume of the cone:
[tex]\[
V_{\text{unfilled}} = V_{\text{cylinder}} - V_{\text{cone}}
\][/tex]
[tex]\[
V_{\text{unfilled}} = 960 \pi - 320 \pi = 640 \pi \quad \text{cubic units}
\][/tex]
The correct answer is:
[tex]\[ 640 \pi \text{ cubic units} \][/tex]
