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Calculus · real student question

Evaluate the triple integral of -2y dz dy dx where x runs from -3 to 3, y from 0 to the square root of 9 minus x squared, and z from 0 to 9 minus x squared minus y squared. Give an exact answer.

Question

Evaluate

∫−33∫09−x2∫0 9−x2−y2(−2y) dz dy dx\int_{-3}^{3}\int_{0}^{\sqrt{9-x^2}}\int_{0}^{\,9-x^2-y^2} (-2y)\,dz\,dy\,dx

Enter an exact answer.

Step-by-step solution

  1. Read the limits as a shape before integrating. The xx and yy limits describe x2+y2≤9x^2+y^2\le 9 with y≥0y\ge 0 — the upper half of the disk of radius 33 — and zz runs from the plane z=0z=0 up to the paraboloid z=9−x2−y2z=9-x^2-y^2. Recognising the half-disk is what makes the cylindrical cross-check in the last step possible.

  2. Integrate in zz first, because the integrand has no zz. The inner integral is just the integrand times the height of the solid:

    ∫09−x2−y2(−2y) dz=−2y (9−x2−y2)\int_{0}^{9-x^2-y^2}(-2y)\,dz=-2y\,(9-x^2-y^2)

  3. Integrate in yy and watch the two terms merge. Expanding, −2y(9−x2)+2y3-2y(9-x^2)+2y^3, so with R2=9−x2R^2=9-x^2:

    [−(9−x2)y2+12y4]09−x2=−(9−x2)2+12(9−x2)2=−12(9−x2)2\Bigl[-(9-x^2)y^2+\tfrac12 y^4\Bigr]_{0}^{\sqrt{9-x^2}}=-(9-x^2)^2+\tfrac12(9-x^2)^2=-\tfrac12(9-x^2)^2

    The upper limit squares to exactly 9−x29-x^2, which is why the two powers of RR combine so cleanly.

  4. Integrate in xx using the evenness of the integrand. Since (9−x2)2=81−18x2+x4(9-x^2)^2=81-18x^2+x^4 is even,

    −12∫−33 ⁣ ⁣(81−18x2+x4)dx=−∫03 ⁣ ⁣(81−18x2+x4)dx-\frac12\int_{-3}^{3}\!\!\bigl(81-18x^2+x^4\bigr)dx=-\int_{0}^{3}\!\!\bigl(81-18x^2+x^4\bigr)dx
    =−[81x−6x3+x55]03=−(243−162+2435)=−6485=-\Bigl[81x-6x^3+\tfrac{x^5}{5}\Bigr]_0^3=-\left(243-162+\frac{243}{5}\right)=-\frac{648}{5}

  5. Confirm the value in cylindrical coordinates. With y=rsin⁡θy=r\sin\theta, dV=r dz dr dθdV=r\,dz\,dr\,d\theta, the same solid is 0≤r≤30\le r\le 3, 0≤θ≤π0\le\theta\le\pi, 0≤z≤9−r20\le z\le 9-r^2:

    ∫0π ⁣ ⁣∫03(−2rsin⁡θ)(9−r2) r dr dθ=−2(∫0π ⁣sin⁡θ dθ)(∫03 ⁣r2(9−r2) dr)\int_0^{\pi}\!\!\int_0^{3}(-2r\sin\theta)(9-r^2)\,r\,dr\,d\theta=-2\left(\int_0^{\pi}\!\sin\theta\,d\theta\right)\left(\int_0^{3}\!r^2(9-r^2)\,dr\right)

    The two factors are 22 and 81−2435=162581-\tfrac{243}{5}=\tfrac{162}{5}, giving −2⋅2⋅1625=−6485-2\cdot 2\cdot\tfrac{162}{5}=-\tfrac{648}{5} — the same exact value from a completely different route.

  6. Interpret the sign. The whole solid sits in y≥0y\ge 0 and the integrand −2y-2y is ≤0\le 0 there, so a negative answer is expected; a positive result would immediately signal a sign slip.

Answer

−6485-\frac{648}{5}

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