Height of a cone given its radius

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this is my first time posting. I have tried this solving this question but am unable to find a solution. We are supposed to use integration. Please help me understand what I am doing wrong? Thank you.

Question: An ice cream cone has radius 2.5 cm at the top. If a scoop of ice cream in the form of a sphere with radius 4 cm is placed on top of the cone, what percentage of the ice cream is outside the cone?

Height of the cone is not provided in the original question, so I am stuck on how to solve this.

Volume of ice cream $= \frac43π4^3 = 256π/3$

Volume of cone $= π6.25\frac h3$

Percentage of ice cream outside of cone = [(Volume of ice cream - Volume of cone)/Volume of ice cream]*100

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Here is your ice cream cone.

enter image description here

Your volume is:

$\int_{-h}^4 (4^2 - x^2) \ dx$

You will need to use the Pythagorean theorem to find $h.$

And I would suggest you think on what $4^2 - x^2$ might actually represent.