Volume of ice cream cone

**Frostking** · May 25th 2009, 08:26 AM

I need to find the volume of an ice cream cone bound by the hemisphere z = (8 = x^2 - y^2)^1/2 and the cone z = (x^2 + y^2)^1/2

I know that is is a double integral but I can't even get started from there. Would someone please push or point me in the right direction? Thanks much for your effort. Frostking

**Spec** · May 25th 2009, 08:44 AM

$\sqrt{x^2+y^2}\leq z \leq \sqrt{8 - x^2 - y^2}$

The projection on the xy-plane, $E$ , is where the two surfaces meet: $\sqrt{x^2+y^2}=\sqrt{8 - x^2 - y^2} \Longleftrightarrow x^2+y^2=4$

So you can calculate it like this: $\iint_E \left( \sqrt{8 - x^2 - y^2}-\sqrt{x^2+y^2}\right)dxdy$

**Spec** · May 25th 2009, 08:52 AM

To understand why we are doing this, consider this: $\int_{\sqrt{x^2+y^2}}^{\sqrt{8 - x^2 - y^2}}\left(\iint_E dxdy\right)dz=\iint_E \left( \sqrt{8 - x^2 - y^2}-\sqrt{x^2+y^2}\right)dxdy$

**Frostking** · May 25th 2009, 08:59 AM

So, if x^2 + y^2 is four then my bounds should be from 0 to 2 for both x and y is this correct? Frostking

**Spec** · May 25th 2009, 09:07 AM

$x^2+y^2=4$ is a circle with the radius 2. So both x and y go from -2 to 2. I always switch to polar coordinates for problems like these though.

Either go with $-\sqrt{4-x^2}\leq y \leq\sqrt{4-x^2},\ -2\leq x\leq 2$ or the other way around $-\sqrt{4-y^2}\leq x \leq\sqrt{4-y^2},\ -2\leq y\leq 2$

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