Trouble with volume/area problems with integrals

• Jul 18th 2008, 11:51 AM
Trouble with volume/area problems with integrals
Okay, so I'm almost finished my webwork assignment, but 3 problems still evade me. I'll say what they are and show what I have done so far.

1. FInd the area of the region enclosed by the curves 2y=3sqrt(x), y=5 and 2y+4x=7.

2. Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=3+(1/x^2), y=3, x=4 and x=8 about the x-axis.

So for 1, I drew a sketch based on what it looked like on my graphing calculator. Nothing too fancy, but just enough so I would know to divide my almost triangular shape into two intervals.

I found the intercepts to be (going from left to right), x=-3/4, 1 and 100/9. I found -3/4 by substituting 5 for y into 2y+4x=7. I found x=1 by setting 3sqrt(x)=7-4x, squaring both sides and solving the quadratic, but I suspect that this may be my mistake. And I found x=100/9 by substituting y=5 into y=(3sqrt(x))/2.

I did it in two parts.
Afinal=A1+a2

A1= integral(-3/4 to 1) (5-((-4x+7)/2))dx

I eventually got 0.8125 for this part.

A2=integral (1 to 100/9) (5-((3sqrt(x))/2)

I got 18.51851852 for this part.

I have no idea what I'm doing wrong, but I'm sure it must be something really little and annoying.
• Jul 19th 2008, 06:22 PM
kalagota
why dont you do it horizontally?

solve for x in terms of y..

(1) $x=\frac{4}{9}y^2$
(2) $x = \frac{7-2y}{4}$

get the intersection of that 2 functions, and that would be $(1,3/2)$

then, $\int_{3/2}^5 \left(\frac{4}{9}y^2 - \frac{7-2y}{4}\right) \, dy$