Consider the transformation T : x = 30 34 u ? 16 34 v , y = 16 34 u + 30 34 v (a) Compute the...

Question:

Consider the transformation {eq}T : x = \frac{30}{34} u - \frac{16}{34}v, \ y = \frac{16}{34}u + \frac{30}{34}v {/eq}

(a) Compute the Jacobian:

{eq}\frac{\partial (x,y)}{\partial (u,v)} = {/eq} _____

(b) The transformation is linear, which implies that it transforms lines into lines. Thus, it transforms the square {eq}S : -34 \leq u \leq 34, \ -34 \leq v \leq 34 {/eq} into a square {eq}T(S) {/eq} with vertices:

{eq}T(34, 34) = {/eq} _____

{eq}T(-34, 34) = {/eq} _____

{eq}T(-34,-34) = {/eq} _____

{eq}T(34,-34) = {/eq} _____

(c) Use the transformation {eq}T {/eq} to evaluate the integral {eq}\iint_{T(S)} x^2 + y^2 dA {/eq}.

Change of Variables:

When we have a region that is difficult to describe in the given coordinates, we can always make a change of variables to simplify things. In this problem we will explore this strategy. We will change from {eq}x,y {/eq} to {eq}u,v {/eq} and apply the following:

{eq}\begin{align*} \iint_R f (x,y) \ dA &= \iint_S f (x(u,v),y(u,v)) \ \bigg| \frac{\partial (x,y)}{\partial (u,v)} \bigg|\ du\ dv \end{align*} {/eq}

where {eq}\frac{\partial (x,y)}{\partial (u,v)} {/eq} is the Jacobian:

{eq}\begin{align*} \frac{\partial(x,y)}{\partial(u,v)} &= \left| \begin{matrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\\\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{matrix} \right| \\\\ &= \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u} \end{align*} {/eq}

Answer and Explanation:

Part A

The Jacobian is

{eq}\begin{align*} \frac{\partial(x,y)}{\partial(u,v)} &= \left| \begin{matrix} \dfrac{\partial }{\partial u} \left( \frac{30}{34} u - \frac{16}{34}v \right) & \dfrac{\partial }{\partial v} \left( \frac{30}{34} u - \frac{16}{34}v \right) \\\\ \dfrac{\partial }{\partial u} \left( \frac{16}{34}u + \frac{30}{34}v \right) & \dfrac{\partial }{\partial v} \left( \frac{16}{34}u + \frac{30}{34}v \right) \end{matrix} \right| \\\\ &= \left| \begin{matrix} \dfrac{15}{17} & -\dfrac{8 }{17} \\\\ \dfrac{8 }{17} & \dfrac{15}{17} \end{matrix} \right| \\\\ &= \frac{225}{289} + \frac{64}{289} \\ &= 1 \end{align*} {/eq}

which is very nice.


Part B

Now we see why they didn't simplify those fractions: so we could see exactly what we need.

{eq}\begin{align*} T(34,34) &= (30 -16, 16+30) = (14, 46) \\ T(-34,34) &= ( -30-16, -16 + 30) = (-46, 14) \\ T(-34,-34) &= ( -30 + 16, -16 - 30) = ( 14, -46) \\ T(34,-34) &= ( 30 + 16, 16 - 30) = (46, -14) \end{align*} {/eq}


Part C

Note that

{eq}\begin{align*} x^2 + y^2 &= \left( \frac{30}{34} u - \frac{16}{34}v \right)^2 + \left( \frac{16}{34}u + \frac{30}{34}v \right)^2 \\ &= \frac{15^2}{17^2} u^2 - \frac{2(30)(16)}{34^2} + \frac{8^2}{17^2} v^2 + \frac{8^2}{17^2} u^2 + \frac{2(30)(16)}{34^2} + \frac{15^2}{17^2} v^2 \\ &= u^2 + v^2 \end{align*} {/eq}

And so we find

{eq}\begin{align*} \iint_R x^2 + y^2 \ dA &= \int_{-34}^{34} \int_{-34}^{34} u^2 + v^2\ du\ dv \\ &= \int_{-34}^{34} \left[ \frac13 u^3 + uv^2 \right]_{-34}^{34} \ dv \\ &= \int_{-34}^{34} \frac13 \left (34^3 - (-34)^3 \right ) + (34+34)v^2 \\ &= \int_{-34}^{34} \frac{78608}3 + 68v^2\ dv \\ &= \left[ \frac{78608}3 v +\frac{68}{3}v^3\right]_{-34}^{34} \\ &= \frac{78608}3 (34+34) + \frac{68}{3}(34^3 - (-34)^3) \\ &= \frac{10690688}{3} \\ &\approx 3.5636 \times 10^6 \end{align*} {/eq}


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Double Integrals: Applications & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 14
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