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A cylinder is to be cut out of the circular face of a solid hemisphere - AQA - A-Level Maths Mechanics - Question 9 - 2020 - Paper 2
Question 9
A cylinder is to be cut out of the circular face of a solid hemisphere. The cylinder and the hemisphere have the same axis of symmetry. The cylinder has height $h$... show full transcript
Worked Solution & Example Answer:A cylinder is to be cut out of the circular face of a solid hemisphere - AQA - A-Level Maths Mechanics - Question 9 - 2020 - Paper 2
Step 1
Show that the volume, V, of the cylinder is given by
Answer
To determine the volume of the cylinder, we start with the formula for the volume of a cylinder:
The base area of the cylinder, which is a circle, can be expressed as:
pi r^2$$ Here, the radius $r$ of the cylinder can be related to the radius $R$ of the hemisphere using the right triangle formed by the radius of the hemisphere, the radius of the cylinder, and the height $h$ of the cylinder: $$R^2 = r^2 + h^2$$ From here, we can express $r$ in terms of $R$ and $h$: $$r = ext{sqrt}(R^2 - h^2)$$ Substituting this back into the volume formula gives us: $$V = pi (R^2 - h^2) h = pi R^2 h - pi h^3$$ This shows that the volume of the cylinder is correctly given by $V = pi R^2 h - pi h^3$.Step 2
Find the maximum volume of the cylinder in terms of R. Fully justify your answer.
Answer
To find the maximum volume of the cylinder, we will first differentiate the volume function with respect to :
pi R^2 h - pi h^3$$ differentiating gives: $$\frac{dV}{dh} = pi R^2 - 3 pi h^2$$ Setting the derivative to zero for maximum volume: $$0 = pi R^2 - 3 pi h^2$$ This simplifies to: $$R^2 = 3h^2$$ Thus, we can express $h$ in terms of $R$: $$h = \frac{R}{\sqrt{3}}$$ Next, we substitute this value of $h$ back into the volume equation to find the maximum volume: $$V = pi R^2 \left( \frac{R}{\sqrt{3}} \right) - pi \left( \frac{R}{\sqrt{3}} \right)^3$$ $$V = \frac{\pi R^3}{\sqrt{3}} - \frac{\pi R^3}{3 \sqrt{3}}$$ Combining these terms gives: $$V = \frac{\pi R^3}{\sqrt{3}} \left( 1 - \frac{1}{3} \right) = \frac{\pi R^3}{\sqrt{3}} \cdot \frac{2}{3} = \frac{2\pi R^3}{3\sqrt{3}}$$ This is the maximum volume of the cylinder in terms of $R$.