**Please help with ****Calculus Problem. I know its kind of tricky but please help?**

* *

*Consider the solid obtained by rotating the region bounded by the given curves about the x-axis*

*y= 4 times sqrt of (64-x^3) y=0, x=5, x=6*

*Find the volume V of this solid
SORRY I MEANT x^2 not x^3 *

* *

Use the Washer Method to solve for the volume of x-axis rotation.

Washer Method = ∫ π F(x)^2 – π f(x)^2 dx

F(x) = 4√(64 – x^2)

f(x) = nothing = 0

∫ π F(x)^2 – π f(x)^2 dx

= ∫ π[4√(64 – x^2)]^2 – π(0)^2 dx

= ∫ π[4√(64 – x^2)]^2 dx

= ∫ π[4√(64 – x^2)]^2 dx

= ∫ π(1024 – 16x^2) dx

= ∫ 1024π dx – ∫ 16πx^2 dx

= 1024πx – 16π ∫ x^2 dx

= 1024πx – 16π ∫ x^2 dx

1024πx – (16πx^3)/3 | From 5 to 6

= [1024π(6) – (16π(6)^3)/3] – [1024π(5) – (16π(5)^3)/3]

= 1616π/3 units squared

Final answer: 1616π/3 units squared or 1692.28 units squared

**It’s Tricky Calculus.wmv**