Volume of the solid obtained by rotating R around the line y=x

In summary, the volume of the solid obtained by rotating $R$ around the line $y = x$ is $\pi\!\!\int_1^3\left(\frac{t-1}{\sqrt2} - \frac{\sqrt2}{\ln3}\ln t\right)^{\!\!2}\frac{dt}{\sqrt2}$.
  • #1
lfdahl
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Let $R$ be the region $\left\{(x, y) : 0 \leq x \leq 1, 3^x − x − 1 \leq y \leq x\right\}$.

Find the volume of the solid obtained by rotating $R$ around the line $y = x$.
 
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  • #2
lfdahl said:
Let $R$ be the region $\left\{(x, y) : 0 \leq x \leq 1, 3^x − x − 1 \leq y \leq x\right\}$.

Find the volume of the solid obtained by rotating $R$ around the line $y = x$.
[sp]Rotate the axes through $45^\circ$ by introducing new coordinates $u = \dfrac{x+y}{\sqrt2}$, $v = \dfrac{y-x}{\sqrt2}$. The line $y=x$ then becomes the $u$-axis.

If $y = 3^x − x − 1$ then $x+y+1 = 3^x$, and so $\ln(x+y+1) = x\ln3$. In terms of $u$ and $v$ that becomes $\ln(\sqrt2u + 1) = \dfrac{\ln3}{\sqrt2}(u-v)$. Solve that for $v$ to get $$v = u - \frac{\sqrt2}{\ln3}\ln(\sqrt2u + 1).$$

The volume of the solid obtained by rotating $R$ around the $u$-axis is then $$V = \pi\!\!\int_0^{\sqrt2}v^2\,du = \pi\!\!\int_0^{\sqrt2}\left(u - \frac{\sqrt2}{\ln3}\ln(\sqrt2u + 1)\right)^{\!\!2}du.$$ Substitute $t = \sqrt2u+1$: $$V = \pi\!\!\int_1^3\left(\frac{t-1}{\sqrt2} - \frac{\sqrt2}{\ln3}\ln t\right)^{\!\!2}\frac{dt}{\sqrt2} = \frac\pi{\sqrt2}\int_1^3\left(\frac12(t-1)^2 - \frac2{\ln3}(t-1)\ln t + \frac2{(\ln3)^2}(\ln t)^2\right)dt.$$

Dealing with the three terms in that last integral, the first one is easy, the other two require integration by parts: $$\int_1^3 (t-1)^2dt = \Bigl[\frac13(t-1)^3\Bigr]_1^3 = \frac83;$$ $$ \int_1^3 (t-1)\ln t\,dt = \Bigl[\frac12(t-1)^2\ln t \Bigr]_1^3 - \int_1^3 \frac12(t^2 - 2t + 1)\frac1t\,dt = 2\ln3 - \Bigl[\frac14t^2 - t + \frac12\ln t\Bigr]_1^3 = 2\ln 3 - \Bigl(-\frac34 + \frac12\ln3 + \frac34\Bigr) = \frac32\ln3; $$ $$ \int_1^3(\ln t)^2dt = \Bigl[t(\ln t)^2 \Bigr]_1^3 - \int_1^3t\frac{2\ln t}tdt = 3(\ln3)^2 - 2\Bigl[t\ln t - t\Bigr]_1^3 = 3(\ln3)^2 - 6\ln3 + 4.$$

Put those results into the integral for $V$ to get $$V = \frac\pi{\sqrt2}\left(\frac43 - 3 + 6 - \frac{12}{\ln3} + \frac8{(\ln3)^2}\right) = \frac\pi{\sqrt2}\left(\frac{13}3 - \frac{12}{\ln3} + \frac8{(\ln3)^2}\right).$$

Numerically, that gives $V\approx 0.086$, which looks about the right order of magnitude from the appearance of a graph of the function.
[/sp]
 
  • #3
Thankyou, Opalg, for a nice and thorough solution! (Yes)
 

1. What is the formula for finding the volume of the solid obtained by rotating R around the line y=x?

The formula for finding the volume of the solid obtained by rotating R around the line y=x is V = π∫ab(f(x))2dx, where a and b are the limits of rotation and f(x) is the function defining the curve R.

2. How do you determine the limits of integration for finding the volume of the solid?

The limits of integration can be determined by finding the points of intersection between the curve R and the line y=x. These points will be the limits of rotation in the formula V = π∫ab(f(x))2dx.

3. Can the volume of the solid obtained by rotating R around the line y=x be negative?

No, the volume of a solid cannot be negative. The formula V = π∫ab(f(x))2dx calculates the absolute value of the volume.

4. How does the shape of the curve R affect the volume of the solid obtained by rotating around the line y=x?

The shape of the curve R will affect the volume of the solid obtained by rotating around the line y=x. If the curve R is symmetric about the line y=x, then the volume will be the same on either side of the line. If the curve R is not symmetric, then the volume will be different on either side of the line.

5. Can the volume of the solid obtained by rotating R around the line y=x be calculated using different units?

Yes, the volume can be calculated using different units as long as the units for the limits of rotation and the function f(x) are consistent. For example, if the limits of rotation are in meters and the function f(x) is in meters, then the volume will be in cubic meters.

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