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Math Help - Volume and Surface Area of a Rotated Solid and Arclength

  1. #1
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    Volume and Surface Area of a Rotated Solid and Arclength

    Hello, I'm back again. I have three problems (two with two parts) that, if you guys wouldn't mind, I would like to have checked. If anything is wrong, I would highly appreciate it if you showed me where I went wrong and how to do it correctly. Thanks infinitely in advance.

    Problem 1.
    Let R be the region bounded by the semicircle y = \sqrt{r^2 - x^2} and the x-axis.

    a. Derive the formula for the volume of a sphere of radius r by revolving R about the x-axis using the disk method or the shell method.

    My Solution (Using Disk Method)


    V = \pi\int^r_0 r^2 dx = \pi[xr^2]^r_0 = \pi r^3

    b. Derive the formula for the surface area of a sphere of radius r by revolving R about the x-axis.

    My Solution (I don't know how to integrate this, please help!)

    A = 2\pi\int^r_0[\sqrt{r^2 - x^2} \sqrt{1 - (\frac{x}{\sqrt{r^2 - x^2}})^2}] dx = ???

    Problem 2. Let R be the region bounded by the lines x = 0, x = h, y = 0, and y = \frac{r}{h}x, where r and h are positive constants.

    a. Derive the formula for the volume of a right circular cone of radius r and altitude h by revolving R about the x-axis using the disk method or the shell method.

    My Solution (Using Disk Method)

    V = \pi\int^h_0(\frac{r}{h}x)^2 dx = \pi\int^h_0\frac{r^2}{h^2}x^2 dx = \pi[\frac{r^2}{3h^2}x^3]^h_0 = \frac{\pi(r^2)h}{3}

    b. Now derive the formula for the surface area of such a cone, again, by revolving R about the x-axis.

    My Solution

    A = 2\pi\int^h_0(\frac{r}{h}x)\sqrt{1 + x^2} dx = 2\pi\int^h_0(\frac{r}{h}x)(1 + x^2)^\frac{1}{2} dx = 2\pi[\frac{r}{3h}(1 + x^2)^\frac{3}{2}]^h_0 = 2\pi[\frac{r}{3h}(1 + h^2)^\frac{3}{2}] - \frac{2\pi(r)}{3h} = \frac{2\pi(r)}{3h}(1 + h^2)^\frac{3}{2} - \frac{2\pi(r)}{3h}

    Problem 3. Calculate the length along the curve y = 1 + x^\frac{3}{2} from (0,1) to (4,9).

    My Solution

    L = \int^4_0 \sqrt{1+(\frac{3}{2}\sqrt{x})^2} dx = \int^4_0\sqrt{1 + \frac{9}{4}x} dx = \int^4_0(1 + \frac{9}{4}x)^\frac{1}{2} dx = [\frac{2}{3}(1 + \frac{9}{4}x)^\frac{3}{2}(\frac{4}{9})]^4_0 = [\frac{8}{27}(1 + \frac{9}{4}x)^\frac{3}{2}]^4_0 = 8 - \frac{8}{27} = \frac{208}{27}

    Once again, thank you.
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  2. #2
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    Hello mturner07
    Quote Originally Posted by mturner07 View Post
    Hello, I'm back again. I have three problems (two with two parts) that, if you guys wouldn't mind, I would like to have checked. If anything is wrong, I would highly appreciate it if you showed me where I went wrong and how to do it correctly. Thanks infinitely in advance.

    Problem 1.
    Let R be the region bounded by the semicircle y = \sqrt{r^2 - x^2} and the x-axis.

    a. Derive the formula for the volume of a sphere of radius r by revolving R about the x-axis using the disk method or the shell method.

    My Solution (Using Disk Method)


    V = \pi\int^r_0 r^2 dx = \pi[xr^2]^r_0 = \pi r^3
    Thanks for showing us your working, but there are a few problems. Here's how you should start the first one.

    The semicircle lies between x = -r and x = r. When we divide the sphere into vertical discs, each disc has a radius y and thickness \delta x. So the total volume is:
    \pi\int_{-r}^ry^2\;dx =\pi\int_{-r}^r(r^2-x^2)\;dx
    Do you want to have another attempt from here?

    Grandad
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  3. #3
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    Hello mturner07

    Here are some more suggestions.
    Quote Originally Posted by mturner07 View Post
    ...Problem 1. Let R be the region bounded by the semicircle y = \sqrt{r^2 - x^2} and the x-axis.
    ...
    b. Derive the formula for the surface area of a sphere of radius r by revolving R about the x-axis.

    My Solution (I don't know how to integrate this, please help!)

    A = 2\pi\int^r_0[\sqrt{r^2 - x^2} \sqrt{1 - (\frac{x}{\sqrt{r^2 - x^2}})^2}] dx = ???
    Your integral is mostly correct: but it should be a + sign, and you have the wrong limits again. Like this:
    A = 2\pi\int_{-r}^ry\sqrt{1+\left(\frac{dy}{dx}\right)^2}\;dx
    Now we have y = \sqrt{r^2-x^2} and \frac{dy}{dx}=-\frac{x}{y}, and you'll find it's easier to write the integral like this before making the substitution:
    \Rightarrow A = 2\pi\int_{-r}^r\sqrt{y^2+y^2\left(\frac{dy}{dx}\right)^2}\;dx
    =2\pi\int_{-r}^r\sqrt{y^2+y^2\left(-\frac{x}{y}\right)^2}\;dx

    =2\pi\int_{-r}^r\left(\sqrt{y^2+x^2}\right)^2\;dx

    =2\pi\int_{-r}^r\left(\sqrt{r^2}\right)^2\;dx
    I'm sure you can finish off from here.
    Problem 2. Let R be the region bounded by the lines x = 0, x = h, y = 0, and y = \frac{r}{h}x, where r and h are positive constants.

    a. Derive the formula for the volume of a right circular cone of radius r and altitude h by revolving R about the x-axis using the disk method or the shell method.

    My Solution (Using Disk Method)

    V = \pi\int^h_0(\frac{r}{h}x)^2 dx = \pi\int^h_0\frac{r^2}{h^2}x^2 dx = \pi[\frac{r^2}{3h^2}x^3]^h_0 = \frac{\pi(r^2)h}{3}
    Correct! Well done!

    b. Now derive the formula for the surface area of such a cone, again, by revolving R about the x-axis.

    My Solution

    A = 2\pi\int^h_0(\frac{r}{h}x)\sqrt{1 + x^2} dx = 2\pi\int^h_0(\frac{r}{h}x)(1 + x^2)^\frac{1}{2} dx = 2\pi[\frac{r}{3h}(1 + x^2)^\frac{3}{2}]^h_0 = 2\pi[\frac{r}{3h}(1 + h^2)^\frac{3}{2}] - \frac{2\pi(r)}{3h} = \frac{2\pi(r)}{3h}(1 + h^2)^\frac{3}{2} - \frac{2\pi(r)}{3h}
    Your initial integral is wrong. \frac{dy}{dx} = \frac{r}{h}, so it should be:
    A = 2\pi\int_0^h\frac{rx}{h}\sqrt{1+\frac{r^2}{h^2}}\;  dx
    Most of the stuff inside the integral is constant, so it just ends up as:
    A = \frac{2\pi r}{h}\sqrt{1+\frac{r^2}{h^2}}\int_0^hx\;dx
    You can finish from here.
    Problem 3. Calculate the length along the curve y = 1 + x^\frac{3}{2} from (0,1) to (4,9).

    My Solution

    L = \int^4_0 \sqrt{1+(\frac{3}{2}\sqrt{x})^2} dx = \int^4_0\sqrt{1 + \frac{9}{4}x} dx = \int^4_0(1 + \frac{9}{4}x)^\frac{1}{2} dx = [\frac{2}{3}(1 + \frac{9}{4}x)^\frac{3}{2}(\frac{4}{9})]^4_0 = [\frac{8}{27}(1 + \frac{9}{4}x)^\frac{3}{2}]^4_0 = 8 - \frac{8}{27} = \frac{208}{27}
    I think this is nearly all correct, but you've got the arithmetic wrong at the end.
    \left[\frac{8}{27}\left(1 + \frac{9}{4}x\right)^\frac{3}{2}\right]^4_0=\frac{8}{27}\left(10^{\frac32}-1\right)
    Grandad
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  4. #4
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    Thanks for the help!

    With your help I get \frac{4}{3}\pi(r)^3 for problem 1a., \pi(r)h\sqrt{1 + \frac{r^2}{h^2}} for problem 2b., and \frac{80\sqrt{10} - 8}{27} for problem 3.

    However, I have a question about your solution to problem 1a. Where does the square that squares \sqrt{y^2 + x^2} come from in the step that has 2\pi\int^r_{-r}(\sqrt{y^2 + x^2})^2 dx?

    Once again, thank you!
    Last edited by mturner07; January 28th 2010 at 12:53 PM. Reason: Added in more specific details.
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  5. #5
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    Hello mturner07
    Quote Originally Posted by mturner07 View Post
    Thanks for the help!

    With your help I get \frac{4}{3}\pi(r)^3 for problem 1a., \pi(r)h\sqrt{1 + \frac{r^2}{h^2}} for problem 2b., and \frac{80\sqrt{10} - 8}{27} for problem 3.

    However, I have a question about your solution to problem 1a. Where does the square that squares \sqrt{y^2 + x^2} come from in the step that has 2\pi\int^r_{-r}(\sqrt{y^2 + x^2})^2 dx?

    Once again, thank you!
    Sorry, that's a typo - it shouldn't be there! Should read
    2\pi\int_{-r}^r\left(\sqrt{y^2+x^2}\right)\;dx
    Grandad
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  6. #6
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    Alright. For problem 1a I get \frac{4}{3}\pi(r)^3, for problem 1b I get 4\pi(r)^2, for problem 2b I get \pi(r)h\sqrt{1 + \frac{r^2}{h^2}}, and for problem 3 I get \frac{80\sqrt{10} - 8}{27}. Are these solutions consistent with yours? Thanks in advance.
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  7. #7
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    Hello mturner07
    Quote Originally Posted by mturner07 View Post
    Alright. For problem 1a I get \frac{4}{3}\pi(r)^3, for problem 1b I get 4\pi(r)^2, for problem 2b I get \pi(r)h\sqrt{1 + \frac{r^2}{h^2}}, and for problem 3 I get \frac{80\sqrt{10} - 8}{27}. Are these solutions consistent with yours? Thanks in advance.
    Yes, I agree with all these answers. The volume and surface area of a sphere are \tfrac43\pi r^3 and 4\pi r^2 respectively.

    You also have a correct expression for the curved surface area of a cone, which can then be simplified as follows:
    \pi rh\sqrt{1 + \frac{r^2}{h^2}}=\pi r\sqrt{h^2+r^2}
    =\pi rl
    where l is the 'slant-height' of the cone; i.e. the distance from a point on the circumference of the base to the vertex.

    Your answer for problem 3 also agrees with mine:
    \frac{8}{27}\left(10^{\frac32}-1\right) = \frac{8(10\sqrt{10}-1)}{27}
    =\frac{80\sqrt{10}-8}{27}
    Grandad
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  8. #8
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    Thank you, I really appreciate all of your help.
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