The base of the solid is a region in the x-y plane bounded by the functions y=x^2 and (1/3)x^2+24. Vertical cross-sections of the solid perpendicular to the x-axis are rectangles 5 units in height. Find the volume of the solid.

Question

amistre64 · Answer

base area times height

amistre64 · Answer

$$\left|pi\int_{a}^{b}(x^2-(\frac{1}{3}x^2+24)dx\right|$$
$$\left|pi\int_{a}^{b}(x^2-\frac{1}{3}x^2-24)dx\right|$$
$$\left|pi(\frac{1}{3}x^3-\frac{1}{9}x^3-24x)\right|^{b}_{a}$$
is your base area

amistre64 · Answer

b is where they cross on the right, and a is the negative of it on the left|dw:1316188210113:dw|

amistre64 · Answer

the absolute value bars might not be needed; you would get a negative result of you by chance subtract the lower from the upper and then instead of ignoring the sign like a rational person, people would start to pull out their hair in disbelief that such a thing could occur

amistre64 · Answer

$$2*\left|pi(\frac{1}{3}x^3-\frac{1}{9}x^3-24x)\right|^{b}_{0}$$

this might be a better or easier solution so you only have to figure out one side and the double it