Finding Volume of Solid Rotated Using Washers and Disks

[twoski]

Homework Statement

Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and y = 4 about the line y = 4

The Attempt at a Solution

Since y=4 is parallel to the x-axis i assume i should be using the 'washer' method.

The points of intersection are x = 2, x = -2... But when i compute the volume i get the wrong answer.

$V = \pi \int ^{2}_{-2} ( x^{4} - 8 )dx = 1/5 x^{5} - 8x | ^{2}_{-2}$

which results in pi * 64/5 which is apparently wrong...

 

[LCKurtz]

Homework Statement

Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and y = 4 about the line y = 4

The Attempt at a Solution

Since y=4 is parallel to the x-axis i assume i should be using the 'washer' method.

The points of intersection are x = 2, x = -2... But when i compute the volume i get the wrong answer.

$V = \pi \int ^{2}_{-2} ( x^{2} - 4 )dx = 1/3 x^{3} - 4x | ^{2}_{-2}$

which results in pi * 16/3 which is apparently wrong...

Check your formula for the washer method.

 

[twoski]

edit: Whoops.

 

[Mark44]

Homework Statement

Using disks or washers, find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and y = 4 about the line y = 4

The Attempt at a Solution

Since y=4 is parallel to the x-axis i assume i should be using the 'washer' method.

The points of intersection are x = 2, x = -2... But when i compute the volume i get the wrong answer.

$V = \pi \int ^{2}_{-2} ( x^{2} - 4 )dx = 1/3 x^{3} - 4x | ^{2}_{-2}$

which results in pi * 16/3 which is apparently wrong...

Your integrand is incorrect. The volume of a typical disk (it's not a washer here) is $\pi$radius²*Δx. You are forgetting to square the radius.

 

[twoski]

$V = \pi \int ^{2}_{-2} ( x^{4} - 16 )dx = 1/5 x^{5} - 16x | ^{2}_{-2}$

So evaluating this, i get pi * -256/5, which is wrong... Is there a method for determining the order in which i subtract the two given functions? Since y=4 is above y=x^2 should i be subtracting y=x^2 from it instead?

 

[LCKurtz]

What is the disk radius? What do you get when you square it?

 

[slider142]

$V = \pi \int ^{2}_{-2} ( x^{4} - 16 )dx = 1/5 x^{5} - 16x | ^{2}_{-2}$

So evaluating this, i get pi * -256/5, which is wrong... Is there a method for determining the order in which i subtract the two given functions? Since y=4 is above y=x^2 should i be subtracting y=x^2 from it instead?

Draw a picture of your region, and draw a single approximate disk of depth Δx. What is its radius?

 

[twoski]

What is the disk radius? What do you get when you square it?

Well if i draw out x^2 and y=4, the distance from the lowest point in x^2 to y=4 is 4. And 4^2 is 16 so i guess i don't follow what you're hinting at.

 

[slider142]

Well if i draw out x^2 and y=4, the distance from the lowest point in x^2 to y=4 is 4. And 4^2 is 16 so i guess i don't follow what you're hinting at.

Great, that's the radius at x = 4. However, the radius varies with x. What is the radius r(x) as a function of x ?

 

[LCKurtz]

Well if i draw out x^2 and y=4, the distance from the lowest point in x^2 to y=4 is 4. And 4^2 is 16 so i guess i don't follow what you're hinting at.

The vertical distance between two curves is $y_{upper}-y_{lower}$. What is that in this problem? What is its square? That's what goes in the integrand as $\pi r^2dx$.

 

[twoski]

Okay it's 4-x^2 so i guess my order was wrong, if that's what you're getting at?

I forgot order doesn't matter as long as you take the absolute value of your answer (since my result is 256/5 if i shift things around). So i guess 256/5 * pi is correct then?

Am i supposed to be squaring (4-x^2)^2 = (16 - 8x^2 + x^4) or should i be squaring them individually, like (4^2) - ((x^2)^2)?

 

[Mark44]

Okay it's 4-x^2 so i guess my order was wrong, if that's what you're getting at?

This is just the distance between a point on the line and a point on the curve.

I forgot order doesn't matter as long as you take the absolute value of your answer (since my result is 256/5 if i shift things around). So i guess 256/5 * pi is correct then?

Am i supposed to be squaring (4-x^2)^2 = (16 - 8x^2 + x^4) or should i be squaring them individually, like (4^2) - ((x^2)^2)?

You're supposed to square (4 - x²), which would be (4 - x²)².

 

[LCKurtz]

The vertical distance between two curves is $y_{upper}-y_{lower}$. What is that in this problem? What is its square? That's what goes in the integrand as $\pi r^2dx$.

Am i supposed to be squaring (4-x^2)^2 = (16 - 8x^2 + x^4) or should i be squaring them individually, like (4^2) - ((x^2)^2)?

The answer is in my quoted post.