# Thread: Calc 1 ellipse volume

1. ## Calc 1 ellipse volume

Find the volume of the solid generated by revolving about the x-axis the region bounded by the x-axis and the upper half of the ellipse

x^2/a^2 + y^2/b^2 = 1

(and thus find the volume of a prolate spheroid)

Here a and b are positive constants with a>b

Volume of the solid of revolution?

Okay, so this one is tripping me up a bit... here's what I've done:

I tried to solve for y and got --
y = square root(b^2(1 - x^2/a^2)) or y = b*sqrt(1 - x^2/a^2)

I thought I would need to square the function and ultimately multiply by pi:

pi int_a^b (b^2(1-x^2/a^2) dx and I moved the b^2 to the beginning with pi and integrated to get:
b^2* pi [ x - x^3/3a^2]

so my final answer, keeping the b and a...
b^2*pi[(b-(b^3)/(3a^2))-(a-(a^3)/(3a^2))]

unfortunately this is not the correct answer. any help?? thanks!

2. Here we use the formula $V = \pi \int y^2 dx$

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.$

$\frac{y^2}{b^2} = 1 - \frac{x^2}{a^2}$

$y^2 = b^2 \left(1 - \frac{x^2}{a^2}\right)$
$y^2 = \frac{b^2}{a^2} (a^2 - x^2)$

substitute into V formula. Set the limits to a and -a (x intersections which enclose the upper area).

$V = \pi \int \limits_{-a}^a \frac{b^2}{a^2} (a^2 - x^2) dx$

$V = \pi \frac{b^2}{a^2} \int \limits_{-a}^a (a^2 - x^2) dx$

$V = \pi \frac{b^2}{a^2} \left[a^2 x - \frac{x^3}{3} \right]_{-a}^a$

$V = \pi \frac{b^2}{a^2} \left[ \left(a^2 *a - \frac{a^3}{3} \right) - \left(a^2 *-a - \frac{(-a)^3}{3} \right) \right]$

$V = \pi \frac{b^2}{a^2} \left( \frac{2a^3}{3} + \frac{2a^3}{3} \right)$

$V = \pi \frac{b^2}{a^2} * \frac{4a^3}{3}$

$V = \frac{4 \pi a^3 b^2}{3a^2}$

$V = \frac{4}{3} \pi ab^2$