The region between the graph f(x)=x^(1/2) and the x-axis for x∈[0,16] is revolved about the line y=8. Which integral will determine the volume of the solid that is generated? I got π(x-64)dx, x∈[0,16] using the yA=1/2[(f(x))^2 - (g(x))^2]dx x∈[a,b] equation, but the answer is ∫π[64−(8−x^(1/2))^2]dx x∈[0,16]. No idea where I went wrong, but maybe this isn't the right equation to use?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

The region between the graph f(x)=x^(1/2) and the x-axis for x∈[0,16] is revolved about the line y=8. Which integral will determine the volume of the solid that is generated? I got π(x-64)dx, x∈[0,16] using the yA=1/2[(f(x))^2 - (g(x))^2]dx x∈[a,b] equation, but the answer is ∫π[64−(8−x^(1/2))^2]dx x∈[0,16]. No idea where I went wrong, but maybe this isn't the right equation to use?

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

are we using shells or discs?
shells add the area of rectangles: 2pi r discs add the area of circles: pi r^2
it looks like disc method to me

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

|dw:1442770213833:dw|
This is a problem semi-related to centroids. I assumed it was Disc Method, but either I did something wrong or I'm overthinking everything.
we have a circle of radius 8, that have a center removed from them of a radius of 8-sqrt(x)
pi (8^2 - (8-sqrt(x))^2) is the area of a disc for a given value of x
it looks like you have an error in your application. can you write it out in more detail? what were your thoughts?
washer method is another name for removing a disc out of a disc ...
I was looking at this in terms of centroids. Most of the equations are the same as the ones for volume. I actually have the Washer Method equation down (I just realized PFFT). The Disc Method equation is just V=πr^2. The way your wrote it as π[8^2 - (8-x^(1/2))^2] is because of the rotation around the y=8, right?
yes we have a cylindar, or radius 8 (center at y=8, and edge at y=0) we are gutting out the middle of it, so at an given x value, we are 8, minus sqrt(x) |dw:1442770823342:dw| since sqrt(x) + r = 8
pi(R^2 - r^2) gives us the portion that we are concerned with.
Thank you very much for you answers~!
youre welcome

Not the answer you are looking for?

Search for more explanations.

Ask your own question