I received this question as part of my homework assignment. And honestly have no idea where to start. I'm pretty sure I'm over-thinking this. Here is the question:
"Set up the integral for the volume of a cone with the base radius $R$ and height $H$"
I looked up the the formula for 'volume of cone' and and just plugged in $H$ and $R$ where normally values would be, but it was graded as incorrect. What am I missing here?
On
It seems to me you're looking for a hint rather than a straight answer, so I'll give you one.
Consider the cone lying on the side, so that the at the origin the x-axis pierces the middle of the cone's base and then also goes through the tip of the cone. This way the base lies in the y-z plane. Integrating over $x$ from the base to the tip, what does each slice $dx$ represent? And what is the area of each of those slices, expressed in terms of $x$, knowing that the cross-section of the cone you're looking at is always a triangle similar to a right-angled triangle with the two sides that are not the hypotenuse equaling $R$ and $H$ in length?
You got it incorrect because you didn't do what the question asked. You are supposed to set up the INTEGRAL for the volume of a cone, not simply look up the volume of a cone and plug in values!
Each cross section of a cone is a circle. The radius of each circle is determined by similar triangles. Thus, your integral should be $$\int_0^H \pi (\frac{R}{H}y)^2 dy = \frac{1}{3}\pi R^2 H$$