# Math Help - using integrals to find volume

1. ## using integrals to find volume

so i have these 2 problems and for the first one im pretty sure ive done it compleeetely wrong, the second, i dont even know how to start.

the first one, the directions are find the volume of the solid obtained by rotationg the region bounded by the given curves about the specified line. y= 1/4x^2. y= 5-x^2
so i used the calculator to give me the intercepts of x=-2, 2 and my integration is going to be pi(5-x^2 - 1/4x^2)^2 i bring the pi out in front of the integral and my next step i want to do is foil out that problem, but that is so massive, it cant be right, so im stuck .

this second one is the base of s is an elliptical region with boundary curve 9x^2 + 4y^2 = 36. cross sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.

class was a few days ago, so this isnt the freshest in my brain, especially that second problem, because i cant even remember how to set it up. i appreciate any help i can get

2. Originally Posted by gbux512
the first one, the directions are find the volume of the solid obtained by rotationg the region bounded by the given curves about the specified line. y= 1/4x^2. y= 5-x^2
what is the "specified line" ?

this second one is the base of s is an elliptical region with boundary curve 9x^2 + 4y^2 = 36. cross sections perpendicular to the x-axis are isosceles right triangles with hypotenuse in the base.
note that the area of a right isosceles triangle in terms of the hypotenuse, H , is (H^2)/4 , so ...

hypotenuse length = 2y ... A(x) = y^2 = 9 - (9x^2)/4

$\displaystyle V = 2\int_0^2 A(x) \, dx$

3. doh, specified line is x axis

4. method of washers ...

$\displaystyle V = \pi \int_a^b [R(x)]^2 - [r(x)]^2 \, dx$

$\displaystyle V = \pi \int_{-2}^2 (5-x^2)^2 - \left(\frac{x^2}{4}\right)^2 \, dx$