Recent content by clipzfan611

I got the answer--it is -64pi/3. OT: However, can you have a negative volume in the first quadrant? Because I graphed the original equations y=x and y=64x^1/4. Thanks for your help.

This is a shells problem with respect to y. I know y^5 would come out to 1/6y^6 but the 1/64 thing is confusing me. Edit: Oh you multiply 1/6 and 1/64?

Yeah I got 6/64y^6 for the first part. Is that right?

Homework Statement 2\pi \int_0^8 (\frac{y^5}{64}- y^2)dy The Attempt at a Solution Tried U-sub and anti-differentiation and none of them got the answer it should be. Help?

2\pi \int_0^8 (\frac{y^5}{64}- y^2)dy Also how do you integrate that? I tried u-sub and it didnt work. Anti differntiation keeps giving me 512/3pi which is wrong because I know it has to be equivalent to the answer to my first question--64/3pi Thank you for your help.

Well, you have to swap x and y, don't you? (While keeping the same functions.) So you just have to make it x= instead of y= right? In terms of the actual equation? So, for shells: 2pi(intergral)((64x)1/4-x)dx, x=0..4 For washers: pi(integral)((y4/64)2-y2)dy, y=0..4 ...when you rotate it about...

"I'll leave it to you to figure out where I got the limits of integration." Because its the highest point where they intersect? "Do you see why the limits of integration are the same in both integrals, even though one is with respect to x and the other y?" Because the 2nd one has 2pi so you...

Homework Statement Find the volume of the solid generated when region R is resolved about the x-axis using the method of "washers" AND "shells" with respect to X. Homework Equations Let R be the region enclosed by the graphs of y=(64x)^(1/4) and y=x The Attempt at a Solution My...

I figured because of the multiplication dot that you had to do what's in paranthesis first, but it wasnt subtractable so sorry for the confusion. Thanks for your help, you were really patient.

With the u's. Are you sure that you multiply 2/3 to both, or the difference of the two?

Thank you. I got: 2/3(3(radical 3)-2(radical 2) Does that seem correct?

Homework Statement Integral: Upper limit is pie/3, lower limit is 0. secxtanx(square root of 1+secx)dx Sorry if this is confusing, I don't know how to use the real math symbols. The Attempt at a Solution I realized it is U-substitution and not the FTC. u=1+secx dx=sextanx du when x=0...