Set Up Definite Integrals for Calc Problems: A-E

[thename1000]

Can you answer this..and show me how to do it? I'm studying..thanks!

For each of the following, show every step required to set up the definite integrals that will solve the problems but do not integrate.

a.) Find the length of the curve y=x^(3/2) from x=0 to x=5

b.) A spring has a natural length of 1 m. If a 24 N force is required to keep it stretched 3 meters beyond its natural length, how much work is required to stretch it from 3 m to 4 m?

c.) Find the length of the solid generated by revolving around the y-axis the region bounded by the x-axis and y=3x^2+x^3 from x=0 to x=1.

d.) A tank(shaped like a bowl) is full of water. Given that water weighs 62.5 lb per cubic foot, find the work required to pump the water out of the tank

e.) For the lamina P formed by the region bounded by y=3sqrt(x) {thats 3rd root of x} the x-axis from x=0 to x=8, find the x-coordinate of the centriod.

 

[ptr]

The length of a curve is given by $$ds^{2}=dx^{2}+dy^{2}$$, given by Pythagoras' Theorem. Dividing throughout by $$dx^{2}$$, we have $$ds= \sqrt{1+(\frac{dy}{dx})^2}$$. Integrating both sides, you can find the length, s, of the curve.

Work is force integrated over distance.

Imagine little disks of circles with their centres on the y axis, then string them together to get a solid, so volume of revolution about the y-axis is $$\int^{b}_a {2 \pi y^2}dx$$. Here, the values, y, of the function are the radii of the razor thin disks.

 

[tacman]

Sure, I'd be happy to help you set up these definite integrals. Here are the steps for each problem:

a.) To find the length of the curve y=x^(3/2) from x=0 to x=5, we can use the arc length formula:

L = ∫√(1 + (dy/dx)^2) dx

First, we need to find dy/dx:
dy/dx = (3/2)x^(1/2)

Now, we can plug this into the arc length formula:
L = ∫√(1 + (3/2)^2x) dx

To solve this integral, we can use u-substitution:
Let u = 1 + (3/2)^2x
Then, du = (3/2)^2 dx

Substituting this into the integral, we get:
L = ∫√u * (2/3)^2 du
= ∫(2/3)√u du
= (2/3) * (2/3) * ∫√u du
= (4/9) * ∫u^(1/2) du
= (4/9) * (2/3)u^(3/2) + C
= (8/27)u^(3/2) + C

Now, we can plug back in for u and evaluate the integral:
L = (8/27)(1 + (3/2)^2x)^(3/2) + C

To find the definite integral from x=0 to x=5, we can plug in the values:
L = (8/27)(1 + (3/2)^2(5))^(3/2) - (8/27)(1 + (3/2)^2(0))^(3/2)
= (8/27)(1 + (45/4))^(3/2) - (8/27)(1 + 0)^(3/2)
= (8/27)(49/4)^(3/2) - (8/27)(1)^(3/2)
= (8/27)(7/2)^3 - (8/27)
= 28/9 - 8/27
= 224/27

Therefore, the length of the curve is approximately 8.296 units.

b.)